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CBSE Solutions for Class 11 English

GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

What is the number of significant figures in 1.050 x 104?

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જવાબ : Four


What is one a.m.u. or one ‘u’?

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જવાબ : 1 a.m.u. or 1 u = 1/12 th mass of an atom of carbon 12.


How are 0.5 ml of NaOH differents from 0.5 M of NaOH?

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જવાબ : 0.5 ml of NaOH means 0.5 mole (20.0 g) of NaOH, 0.5M of NaOH means that 0.5 mole (20.0g) of NaOH are dissolved in 1L of its solution.


State Avogadro’s law.

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જવાબ : Equal volumes of all gases under the conditions of same temperature and pressure contain the same number of molecules.


Name two factors that introduce uncertainty into measured figures.

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જવાબ : (i) Reliability of measuring instrument.
(ii) Skill of the person making the measurement.


How many oxygen atoms are there in 18 g of water? 

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જવાબ : Molar mass of water is 18 g/mol.
Number of oxygen atoms is 18 g of water = 6.02 x 1023


How is empirical formula of a compound related to its molecular formula?

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જવાબ : Molecular formula = (Empirical formula)n where n is positive integer.


Define law of conservation of mass.

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જવાબ : It states that matter can neither be created nor destroyed.


Give an example of a molecule in which the ratio of the molecular formula is six times the empirical formula.

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જવાબ : The compound is glucose. Its molecular formula is C6H12O6, while empirical formula is CH2O.


What is an atom according to Dalton’s atomic theory?

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જવાબ : According to Dalton’s atomic theory, an atom is the ultimate particle of matter which cannot be further divided.


Why air is not always regarded as homogeneous mixture?

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જવાબ : This is due to the presence of dust particles.


Define the term ‘unit’ of measurement.

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જવાબ : It is defined as the standard of reference chosen to measure a physical quantity.


What is the SI unit of molarity?

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જવાબ : SI unit of molarity = mol dm-3


What do you understand by stoichiometric coefficients in a chemical equation?

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જવાબ : The coefficients of reactant and product involved in a chemical equation represented by the balanced form, are known as stoichiometric coefficients.
For example, N2(g) + 3H2(g) ———–> 2 NH3(g)
The stoichiometric coefficients are 1, 3 and 2 respectively.


If two elements can combine to form more than one compound, the masses of

one element that combine with a fixed mass of the other element, are in whole-number ratio.

(a) Is this statement true?

(b) If yes, according to which law?

(c) Give one example related to this law.

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જવાબ : (a) Yes, the statement is true. (b) According to the law of multiple proportions (c), Hydrogen and oxygen react to form water and hydrogen peroxide H2 + 1/2O2 → H2O H2 + O2 → H2O2 Masses of oxygen which combine the fixed mass of hydrogen are in the ratio 16:32 or 1


45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous

oxide was formed. The reaction is given below:

2N2(g) + O2(g) → 2N2O(g)

Which law is being obeyed in this experiment? Write the statement of the law?

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જવાબ : The above experiment proves Gay-Lussac’s law which states that gases combine or produced in a chemical reaction in a simple whole-number ratio by volume provided that all gases are the same temperature and pressure.


Calculate the mass percent of calcium, phosphorus and oxygen in calcium

phosphate Ca3(PO4)

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જવાબ : Molecular mass of Ca3(PO4) = 3*40+2*31+8*16 =310 Mass per cent of Ca = 3*40/310*100 = 38.71% Mass per cent of P = 2*31/310*100 = 20% Mass per cent of O = 8*16/310 = 41.29%


What is the difference between molality and molarity?

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જવાબ : Molarity is the number of moles of solute dissolved in 1 litre of the solution. Molality is the number of moles of solute present in 1kg of the solvent.


What is the symbol for the SI unit of a mole? How is the mole defined?

 

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જવાબ : The symbol for the SI unit of the mole is mol. A mole is defined as the amount of substance that contains as many entities as there are atoms in 12g carbon.


How many significant figures should be present in the answer to the following

calculations?

2.5 1.25 3.5/2.01

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જવાબ : Two significant figures should be present in this. Since the least number of significant figures from the given figure is 2 (in 2.5 and 3.5).


What will be the mass of one atom of C-12 in grams?

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જવાબ : 1 mole of carbon atom = 12g= 6.022 × 1023 atoms.


Calculate the average atomic mass of hydrogen using the following data :

Isotope %

Natural abundance

Molar mass

1H

99.985

1

2H

0.015

2

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જવાબ : Average atomic mass = 99.985*1+0.015*2/100 =099.985*1+0.015*2/100 =1.00015u


The reactant which is entirely consumed in the reaction is known as limiting reagent.

In the reaction 2A + 4B → 3C + 4D, when 5 moles of A react with 6 moles of B,

then

(i) which is the limiting reagent?

(ii) calculate the amount of C formed?

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જવાબ : (i) B will be the limiting reagent as it gives a lesser amount of product. (ii) Let B is completely consumed 4 mol B gives 3 mol C 6 mol B will give 3/4 *6 mol C =4.5 mol C


The volume of a solution changes with change in temperature, then, will the molality of the solution be affected by temperature? Give a reason for your answer.

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જવાબ : Mass does not change as the temperature changes. Therefore, the molality of a solution does not change. Molality = moles of solute/ weight of solvent (in g) *1000


The density of 3 molal solutions of NaOH is 1.110 g mL–1. Calculate the molarity of the solution.

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જવાબ : 3 molal solution of NaOH = 3 moles of NaOH dissolved in 1000g water 3 mole of NaOH = 3*40g = 120g Density of solution = 1.110gmL-1 Volume = mass/density = 1120g/1.110gmL-1 =1.009L Molarity of the solution = 3/1.009 = 2.97M


Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place.

Zn + 2HCl → ZnCl2 + H2

Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCl. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of Zn = 65.3 u.

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જવાબ : 1 mol of gas occupies = 22.7L Volume at STP atomic mass of Zn = 65.3u From the above equation, 65.3g of Zn when reacts with HCl produces = 22.7L H2 at STP Therefore, 32.65g of Zn when reacts with HCl will produce = 22.7 * 32.65/65.3 =11.35L of H2 at STP


Calculate the molecular mass of the following:
(i) H20 (ii) C0(iii) CH4

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જવાબ : (i) Molecular mass of H2O = 2(1.008 amu) + 16.00 amu=18.016 amu
(ii) Molecular mass of CO2= 12.01 amu + 2 x 16.00 amu = 44.01 amu
(iii) Molecular mass of CH4= 12.01 amu + 4 (1.008 amu) = 16.042 amu


Chlorine is prepared in the laboratory by treating manganese dioxide (Mn02) with aqueous hydrochloric acid according to the reaction.
4 HCl (aq) + Mn02 (s) ———–> 2 H2O (l) + MnCl2(aq) +Cl2(g)

How many grams of HCl react with 5.0 g of manganese dioxide? (Atomic mass of Mn = 55 u)

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જવાબ : 1 mole of Mn02, i.e., 55 + 32 = 87 g Mn0react with 4 moles of HCl, i.e., 4 x 36.5 g = 146 g of HCl


What will be the mass of one  12C atom in g ?

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જવાબ :


How are 0.50 mol Na2C0and 0.50 M Na2C0different?

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જવાબ : Molar mass of Na2C03= 2 x 23 + 12 + 3 x 16 = 106g mol-1 0.50 mol Na2C0means 0.50 x 105 g = 53 g 0. 50 M Na2C03 means 0.50 mol, i.e., 53 g Na2C0are present in 1 litre of the solution


If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour could be produced?

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જવાબ : Hand 02 react according to the equation 2H2(g) + 02 (g) ——>2H2O (g)  Thus, 2 volumes of H2 react with 1 volume of 02 to produce 2 volumes of water vapour. Hence, 10 volumes of H2 will react completely with 5 volumes of 02 to produce 10 volumes of water vapour.


If the speed of light is 3.0 x 108 ms-1, calculate the distance covered by light in 2.00 ns.

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જવાબ :


Round up the following upto three significant figures:
(i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808

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જવાબ : (i) 34.2 (ii) 10.4 (iii) 0.0460 (iv) 2810


How many significant figures are present in the following?
(i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,000
(v) 500.0 (vi) 2.0034

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જવાબ : (i) 2 (ii) 3 (iii) 4 (iv) 3 (v) 4 (vi) 5.


Express the following in scientific notation:
(i) 0.0048 (v) 6.0012 (ii) 234,000 (iii) 8008 (iv) 500.0

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જવાબ :


What do you mean by significant figures?

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જવાબ : The digits in a properly recorded measurement are known as significant figures. It is also defined as follows. The total numbers of figures in a number including the last digit whose value is uncertain is called number of significant figures.


What is the S.I. unit of mass?

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જવાબ :  S.I. unit of mass is kilogram (kg).


Pressure is determined as force per unit area of the surface. The S.I. unit of pressure, pascal, is as shown below:1 Pa = 1 Nm-2.If mass of air at sea level is 1034 g cm-2,calculate the pressure in pascal.

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જવાબ : Pressure is the force (i.e., weight) acting per unit area But weight = mg


If the density of methanol is 0.793 kg L -1, what is its volume needed for making 2.5 L of its 0.25 M solution?

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જવાબ :


What is the concentration of sugar (C12H22O11) in mol L -1 if its 20 g are dissolved in enough water to make a final volume up to 2 L?

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જવાબ :


In three moles of ethane (C2H6), calculate the following:
(i) Number of moles of carbon atoms (ii) Number of moles of hydrogen atoms
(iii) Number of molecules of ethane

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જવાબ :  (i) 1 mole of C2H6 contains 2 moles of carbon atoms
.•. 3 moles of C2H6 will C-atoms = 6 moles
(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms
.•. 3 moles of C2H6 will contain H-atoms = 18 moles


Calculate the atomic mass (average) of chlorine using the following data:

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જવાબ :


Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.8 g mol-1(Atomic mass: Fe = 55.85, O = 16.00 amu)Calculation of Empirical Formula. See Q3.

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જવાબ : Empirical formula mass of Fe20= 2 x 55.85 + 3 x 16.00 = 159.7 g mol1
Hence, molecular formula is same as empirical formula, viz.,Fe203.


How much copper can be obtained from 100 g of copper sulphate (CuSO4 )? (Atomic mass of Cu= 63.5 amu)

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જવાબ : 1 mole of CuS0contains 1 mole (1 g atom) of Cu Molar mass of CuS04= 63.5 + 32 + 4 x 16 = 159.5 g mol-1 Thus, Cu that can be obtained from 159.5 g of CuS0= 63.5 g


Define the law of multiple proportions. Explain it with two examples. How does this law point to the existence of atoms?

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જવાબ : When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other bear a simple ratio to one another is the law of multiple proportions.

For example, carbon combines with oxygen to form two compounds they are carbon dioxide and carbon monoxide

The masses of oxygen which combine with a fixed mass of carbon in carbon dioxide and carbon monoxide are 32 and 16. Therefore oxygen bear: 32:16 ratio or 2:1

Example 2: Sulphur combines with oxygen to form sulphur trioxide and sulphur dioxide

The masses of oxygen which combine with a fixed mass of sulphur in SO3 and SO2 are 48 and 32. Therefore oxygen bear a ratio of 48:32 or 3:2


Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given below:

CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)

What mass of CaCl2 will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3? Name the limiting reagent. Calculate the number of moles of CaCl2 formed in the reaction.

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જવાબ : No: of moles of HCl taken = MV/1000 = 0.76*250/1000 = 0.19

No: of moles of CaCO3 = Mass/Molar mass = 1000/100 = 10

1. When CaCO3 is completely consumed

1 mol of CaCO3 = 1 mol CaCl2

10 mol CaCO3 = 10mol CaCl2

2. When HCl is completely consumed.

2 mol HCl = 1 mol CaCl2

0.19mol HCl = ½  × 0.19mol CaCl2 = 0.095 mol CaCl2

HCl will be the limiting reagent and the number of moles of CaCl2 formed will be 0.095mol


A vessel contains 1.6 g of dioxygen at STP (273.15K, 1 atm pressure). The gas is now transferred to another vessel at a constant temperature, where the pressure becomes half of the original pressure. Calculate

(i) the volume of the new vessel.

(ii) a number of molecules of dioxygen.

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જવાબ : (i) Moles of oxygen = 1.6/32 = 0.05mol

At STP, 1 mol of O2 = 22.4L

Then volume of O2 = 22.4 × 0.05 = 1.12L

V1 = 1.12L

V2 =?

P1 = 1atm

P2 = ½ = 0.5atm

According to Boyle’s law, p1V1 = p2V2

Substituting the values

V2 = 1 × 1.12/0.5 = 2.24L

(ii) No of molecules in 1.6g or 0.005mol = 6.022 × 1023 × 0.05 = 3.011 ×  1022


A box contains some identical red coloured balls, labelled as A, each weighing 2 grams. Another box contains identical blue coloured balls, labelled as B, each weighing 5 grams. Consider the combinations AB, AB2, A2B and A2B3 and show that the law of multiple proportions is applicable.

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જવાબ :

  AB

ab2

A,B

A2B3

Mass of A (in g)

2

2

4

415

Mass of B (in g)

5

10

5

 
Masses of B which combine with a fixed mass of A are

10g, 20g, 5g, 15g

2 : 4 : 1 : 3

This is the simple whole-number ratio.


If 4 g of NaOH dissolves in 36 g of H2O, calculate the mole fraction of each component in the solution. Also, determine the molarity of the solution (specific gravity of solution is 1g mL–1).

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જવાબ : Mole fraction of H2O = No; of moles of H2O/ Total no: of moles (H2O+NaOH)

No: of moles of H2O = 36/18=2moles

No: of moles of NaOH = 4/40=0.1mol

Total no: of moles = 2+0.1= 2.1

Mole fraction of H2O = 2/2.1 = 0.952

Mole fraction of NaOH = 0.1/2.1 = 0.048

Mass of solution = Mass of H2O + Mass of NaOH = 36+4=40G

Volume of the solution = 40/1 = 40mL

Molarity = 0.1/0.04 = 2.5M


Calculate the mass percent of different elements present in sodium sulphate (Na2 SO4).

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જવાબ :


Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

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જવાબ : The balanced equation for the combustion of carbon in dioxygen/air is

(i) In air, combustion is complete. Therefore,C0produced from the combustion of 1 mole of carbon = 44 g.(ii) As only 16 g of dioxygen is available, it can combine only with 0.5 mole of carbon, i.e., dioxygen is the limiting reactant. Hence,C0produced = 22 g.(iii) Here again, dioxygen is the limiting reactant. 16 g of dioxygen can combine only with 0.5 mole of carbon.C02 produced again is equal to 22 g.


Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1

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જવાબ : 0.375 M aqueous solution means that 1000 mL of the solution contain sodium acetate = 0.375 mole


Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41 g mL-1 and the mass percent of nitric acid in it is being 69%.

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જવાબ : Mass percent of 69% means that 100 g of nitric acid solution contain 69 g of nitric acid by mass.
Molar mass of nitric acid HNO3= 1 + 14 + 48 = 63 gmol-1


A sample of drinking water was found to be severely contaminated with chloroform, CHCly supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass
(ii) Determine the molality of chloroform in the water sample.

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જવાબ :


Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:(i) N2 (g) + 3H2(g) —–> 2NH3 (g)
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?

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જવાબ :


In the reaction, A + B2——> AB2, identify the limiting reagent, if any, in the following mixtures
(i) 300 atoms of A + 200 molecules ofB
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules ofB
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B

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જવાબ : (i) According to the given reaction, 1 atom of A reacts with 1 molecule of B
.•. 200 molecules of B will react with 200 atoms of A and 100 atoms of A will be
left unreacted. Hence, B is the limiting reagent while A is the excess reagent.
(ii) According to the given reaction, 1 mol of A reacts with 1 mol of B
.•. 2 mol of A will react with 2 mol of B. Hence, A is the limiting reactant.
(iii) No limiting reagent.
(iv) 2.5 mol of B will react with 2.5 mol of A. Hence, B is the limiting reagent.
(v) 2.5 mol of A will react with 2.5 mol of B. Hence, A is the limiting reagent.


Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040.

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જવાબ :


Calculate:
(a) Mass of 2.5 gram atoms of magnesium,
(b) Gram atom in 1.4 grams of nitrogen (Atomic mass Mg = 24, N = 14)

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જવાબ : (a) 1 gram atom of Mg = 24g
2.5 gram atoms of Mg = 24 x 2.5 = 60g
(b) 1 gram atom of N = 14g;
14g of N = 1 gram atom 1
1.4g of N = 1/14 x 1.4 = 0.1 gram atom.


The density of water at room temperature is 1.0 g/mL. How many molecules are there in a drop of water if its volume is 0.05 mL?
 

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જવાબ :


A flask P contains 0.5 mole of oxygen gas. Another flask Q contains 0.4 mole of ozone gas. Which of the two flasks contains greater number of oxygen atoms?

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જવાબ : 1 molecule of oxygen (O2) = 2 atoms of oxygen
1 molecule of ozone (O3) = 3 atoms of oxygen


The Vapour Density of a gaseous element is 5 times that of oxygen under similar conditions. If the molecule is triatomic, what will be its atomic mass?

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જવાબ : Molecular mass of oxygen = 32 u


How much copper can be obtained from 100 g of copper sulphate (CuSO4)?

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જવાબ : 1 mole of CuSO_{4}CuSO4​ contains 1 mole of Cu.

Molar mass of CuSO_{4}CuSO4​

= (63.5) + (32.00) + 4(16.00)

= 63.5 + 32.00 + 64.00

= 159.5 grams

159.5 grams of CuSO_{4}CuSO4​ contains 63.5 grams of Cu.

Therefore, 100 grams of CuSO_{4}CuSO4​ will contain \frac{63.5\times 100g}{159.5}159.563.5×100g​ of Cu.

= \frac{63.5\times 100}{159.5}159.563.5×100​

=39.81 grams


Explain law of multiple proportions with an example.

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જવાબ : The law of multiple proportions rays that if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of other element are in a ratio of small whole numbers. e.g. hydrogen and oxygen can combine to form water as well as hydrogen peroxide.

Here, the masses of oxygen (16g & 32g) which combine with a fixed mass of hydrogen (2g) bear a simple ratio i.e., 16:32 = 1:2.


Write Postulates of Dalton’s atomic theory.

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જવાબ : Postulates of Dalton’s atomic theory –

1. Matter consists of indivisible atoms.

2. All the atoms of a given element have identical properties including atomic mass. Atoms of different element differ in mass.

3. Compounds are formed when atoms of different elements combine in a fixed ratio.

4. Chemical reaction involves reorganization of atoms. These are neither created nor destroyed


Calculate the number of moles in the following masses –

(i) 7.85g of Fe

(ii) 7.9mg of Ca

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જવાબ :


What volume of 10M HCl and 3M HCl should be mixed to obtain 1L of 6M HCl solution?

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જવાબ : Let the required volume of 10M HCl be V liters.

Then, the required volume of 3M HCl be (1 – V) Liters.

M1V1 + M2V2 = M3V3

Then the volume of 10M HCl required = 428mL

& volume of 3M HCl required = 1000mL – 428mL = 572mL


4 litres of water are added to 2L of 6 molar HCl solutions.What is the molarity of resulting solution?

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જવાબ : Initial volume, V1 = 2L

Final volume, V2 = 4L + 2L = 6L

Initial molarity, M1 = 6M

Final molarity = M2

M1V1 = M2V2

Thus the resulting solution is 2M HCl.


Calculate the weight of lime (CaO) obtained by heating 2000kg of 95% pure lime stone (CaCO3)

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જવાબ :


How much potassium chlorate should be heated to produce 2.24L of oxygen at NTP?

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જવાબ : 2KClO3 à 2KCl + 3O2

2moles 3moles

2(39 + 35.5 + 3 x 16) 22.4 x 3L

= 245g = 67.2L

67.2L of oxygen is produced from 245g of KClO3

= 8.17g of KClO3


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Match the following prefixes with their multiples:

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જવાબ :

Match the following

Physical quantity

Unit

(i) Molarity

(ii) Mole fraction

(iii) Mole

(iv) Molality

(v) Pressure

(vi) Luminous intensity

(vii) Density

(viii) Mass

(a) g mL–1

(b) mol

(c) Pascal

(d) Unitless

(e) mol L–1

(f) Candela

(g) mol kg–1

(h) Nm–1

(i) kg

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જવાબ :

(i → e)

(ii → d)

(iii → b)

(iv → g)

(v → c)

(vi → f)

(vii → a)

(viii → i)

(i) 88 g of CO2

(ii) 6.022 ×1023 molecules of H2O

(iii) 5.6 litres of O2 at STP

(iv) 96 g of O2

(v) 1 mol of any gas

(a) 0.25 mol

(b) 2 mol

(c) 1 mol

(d) 6.022 × 1023 molecules

(e) 3 mol

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જવાબ :

A → b

B → c

C → a

D → e

E → d

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