NCERT Solutions for Class 11 Chemistry Chapter 1 Some basic Concepts of Chemistry

NCERT Solutions For Class 11, Chemistry, Chapter 1, Some Basic Concepts Of Chemistry is very important chapter of Class 11 and also for class 12 and many competitive exams. Chapter 1 class 11 chemistry cover basic fundamentals which are helpful in learning and understanding chemistry. Solutions for chapter 1, class 11 given below.

Table of Contents

Topics included in NCERT Solutions for Class 11 Chemistry Chapter 1 Some basic Concepts of Chemistry:

Topic	Name of the topic
1	Some Basic Concepts of Chemistry
1.1	Importance of Chemistry
1.2	Nature of Matter
1.3	Properties of Matter and their Measurement
1.4	Uncertainty in Measurement
1.5	Laws of Chemical Combinations
1.6	Dalton’s Atomic Theory
1.7	Atomic and Molecular Masses
1.8	Mole Concept and Molar Masses
1.9	Percentage Composition
1.10	Stoichiometry and Stoichiometric Calculations

Class 11 chemistry chapter 1 questions with answers.

Q.1.1 Calculate the molecular mass of the following:

(i) H₂O

(ii) CO₂

(iii) CH₄

Ans: (i) Molecular mass of H₂O = 2 (1.008 amu) + 16.00 amu = 18 016 amu

(ii) Molecular mass of CO₂ = 12.01 amu + 2 (16.00 amu) = 44.01 amu

(iii) Molecular mass of CH₄ = 12.01 amu + 4 (1.008 amu) = 16 .042

Q.1.2: Calculate the mass per cent of different elements present in sodium sulphate (Na₂SO₄).

Ans: Mass per cent can be calculated as:

Mass % of an element = $\frac{Molar\,mass\,\,of\,\,that\,\,element}{Molar\,\,mass\,\,of\,\,the\,\,compound}$x 100

Now, molar mass of Na₂SO₄ = (2 x 23.0) + 32.0 + (4 x 16.0) = 142 g mol^-1

Mass percent of Sodium (Na) = $\frac{46}{142}$×100 = 32.39 %

Mass percent of Sulphur (S) = $\frac{32}{142}$×100 = 22.54 %

Mass percent of Oxygen (O) = $\frac{64}{142}$×100 = 45.07 %

Q.1.3: Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Ans: Given,

Mass % of Fe = 69.9

Mass % of O = 30.1

The empirical formula can be calculated as :

Therefore, Empirical formula = Fe₂O₃.

Q.1.4: 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.

Ans: The balanced equation for the combustion of carbon in air is :

(i) when 1 mole of carbon is burnt in air, 44 g (1mol) of CO₂is produced.

(ii) As only 16 g of dioxygen is available, it can combine only with half mole (0.5 mol) of carbon. Hence, CO₂ produced = ½ mol or 22 g.

(ii) As only 16 g of dioxygen is available, it can combine only with half mole (0.5 mol) of carbon. Hence, CO₂ produced = ½ mol or 22 g. here , O₂is the limiting agent.

Q.1.5: 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.

Ans: Assume that, A = Solute, B = Solvent and (A+B) = Solution

Given, Mass of sodium acetate =?

Molar mass of sodium acetate = 82.0245 g mol^–1

Molarity (M) of the solution = 0.375 M

Volume of the solution (V) = 500 ml

0.375 M aqueous solution means that 1000 mL of the solution contain sodium acetate

= 0.375 mol

∴ 500 mL of the solution should contain sodium acetate = $\frac{0.375}{2}$ mole

Molar mass of sodium acetate given = 82.0245 g mol^–1

As we know that,

mol(n) = $\frac{Given\,\,mass\,\,(w)}{Molar\,mass\,\,(M)}$

Therefore,

Mass of sodium acetate = mol × Molar mass (g mol^-1)

Mass of sodium acetate = $\frac{0.375}{2}$mole × 82.0245 g mol^-1=15.380 g

Q.1.6 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 per cent of nitric acid in it being 69 %.

Ans: Given, Density (d) = 1.41 g mL^-1.

Mass per cent of nitric acid = 69 %

So, 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+(16×3) = 63 g mol^-1

Moles of nitric acid (n) = $\frac{Given\,\,mass}{Molar\,mass}=\frac{69\,g}{63\,\,g\,\,mo{{l}^{-1}}}$= 1.095 mole

As we know that, Density (d)= $\frac{Mass\,\,(m)}{Volume\,(V)}$

Volume of nitric acid solution, V = $\frac{Mass\,\,(m)}{Density\,(d)}=\frac{100\,\,g}{1.41\,\,g\,\,m{{l}^{-1}}}$=70.92 ml or 0.07092 L

Therefore, conc. of HNO₃, in mole/L = $\frac{1.095\,mole}{0.07092\,\,L}$ = 15.44 M

Q.1.7 How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?

Ans: Given, Copper sulphate (CuSO₄) = 100 g

Atomic mass of Cu = 63.5 amu

1 mole of CuSO₄ contains = 1 mole of Cu

Molar mass of CuSO₄ = 63.5 + 32 + (4 x16) = 159.5 g mol^-1

∵ Amount of Cu can be obtained from 159.5 g CuSO₄ = 63.5 g

∴ Amount of Cu can be obtained from 100 g CuSO₄ = $\frac{63.5}{159.5}$x 100 = 39.81 g

Q.1.8: Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.

Ans: Given, Mass per cent of iron (Fe) = 69.9

Mass per cent of iron (O) = 30.1

To calculate molecular formula, first we need to find Empirical formula of Fe₂O₃

The empirical formula can be calculated as :

Therefore, Empirical formula = Fe₂O₃.

Now, calculation of Empirical formula mass of Fe₂O₃

= (2 x 55.85) + (3×16.00) = 159.7 g mol^-1

∵ n = $\frac{Molar\,mass}{E.P.\,mass}=\frac{159.8}{159.7}$ = 1

Molecular formula = n x Empirical formula

Molecular formula = 1 x Fe₂O₃

Hence, Molecular formula = Fe₂O₃ (Same as Empirical formula)

Q.1.9 Calculate the atomic mass (average) of chlorine using the following data:

Ans: The average atomic mass of Chlorine can be calculated as:

Average atomic mass = (0.7577 x 34.9689 amu) + (0.2423 x 36.9659 amu)

= (26.4959 + 8.9568) amu

= 35.4527 amu

Q.1.10 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.

Ans: (i) ∵ 1 mole of C₂H₆ contains moles of C atoms = 2 mole

∴ 3 moles of C₂H₆ contains moles of C atoms = 3 x 2 = 6 mole of C atoms

(ii) ∵ 1 mole of C₂H₆ contains moles of H atoms = 6 mole

∴ 3 moles of C₂H₆ contains moles of H atoms = 3 x 6 = 18 mole of H atoms

(iii) ∵ 1 mole of C₂H₆ contains number of molecules of ethane = 6.023 × 10²³

∴ 3 moles of C₂H₆ contains number of molecules of ethane = 3 X 6.023 × 10²³ = 18.069 X10²³

Q.1.11 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 2L?

Ans:

Q.1.12 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?

Ans: Molar mass of methanol (CH₃OH) = 32 g mol^-1 = 0.032 kg mol^~1

Molarity of the solution, M = $\frac{0.793\,\,kg\,\,{{L}^{-1}}}{0.032\,kg\,\,mo{{l}^{-1}}}$ = 24.78 mol L^-1

Now, applying M₁ x V₁= M₂ x V₂

24.78 x V₁ = 0.25 x 2.5 L

V₁ = 0.02522 L = 25.22 ml

Hence, 25.22 ml of methanol is needed.

Q.1.13 Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: 1Pa = 1N m^–2. If mass of air at sea level is 1034 g cm^–2, calculate the pressure in pascal.

Ans: According to question, pressure is force per unit area of the surface

= $\frac{F}{A}=\frac{mg}{A}$

= $\frac{1034\,g\,\times \,9.8\,m{{s}^{-2}}}{c{{m}^{2}}}\times \frac{1\,kg}{1000\,g}\times \frac{100\,cm\,\times \,100\,cm}{1\,m\times 1\,m}\times \frac{1\,N}{kg\,m\,{{s}^{-2}}}\times \frac{1\,Pa}{1\,N{{m}^{-2}}}$ = 1.01332 x 10 Pa

Q.1.14 What is the SI unit of mass? How is it defined?

Ans: The SI unit of mass is the kilogram (kg), which is a measure of how much matter an object has. We use kilograms to measure everything from the weight of a person to the mass of a planet.

The kilogram is defined as the mass of a specific cylinder made of platinum and iridium, called the International Prototype of the Kilogram (IPK), which is kept in a safe place in France.

The weight of this cylinder is used as a standard to compare the weight of other objects.

Q.1.15 Match the following prefixes with their multiples:

Ans: micro = 10^-6, deca = 10, mega = 10⁶, giga = 10⁹, femto = 10^-15.

Q.1.16 What do you mean by significant figures?

Ans: Significant figures are the digits in a numerical value that are important and carry meaning based on the precision of the measurement. They indicate the level of accuracy or uncertainty in the result.

Q.1.17 A sample of drinking water was found to be severely contaminated with chloroform, CHCl₃, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

(i) Express this in per cent by mass.

(ii) Determine the molality of chloroform in the water sample.

Ans:

Q.1.18 Express the following in the scientific notation:

(i) 0.0048

(ii) 234,000

(iii) 8008

(iv) 500.0

(v) 6.0012

Ans:

(i) 4.8 x10^-3

(ii) 2.34 x10⁵

(iii) 8.008 x 10³

(iv) 5.000 x 10²

(v) 6.0012 x 10⁰

Q.1.19 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

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

Q.1.20 Round up the following up to three significant figures:

(i) 34.216

(ii) 10.4107

(iii) 0.04597

(iv) 2808

Ans:

(i) 34.2

(ii)10.4

(iii) 0.0460

(iv) 2810

Q.1.21: The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement.

(b) Fill in the blanks in the following conversions:

(i) 1 km = ……………mm =……pm

(ii) 1 mg = …………..kg =……….ng

(iii) 1 mL = …………..L =………….dm3

Ans: (a) if we Fix the mass of dinitrogen (N₂) as 28 g, masses of dioxygen combined will be 32, 64, 32 and 80 g in the given four oxides.

These are in the ratio 1:2:1:5 which is a simple whole number ratio. Hence, given data obey the law of multiple proportions.

Law of multiple proportions: “if two elements combine to form more than one compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers”.

Q.1.22 If the speed of light is 3.0 x 10⁸ m s^–1, calculate the distance covered by light in 2.00 ns.

Ans: As we know that,

Speed = $\frac{Distance\,travelled}{Time\,taken}$

Distance = Speed X Time

Distance = (3.0 x 10⁸ m s^–1 ) X 2.00 ns or

= 3.0 ×10⁸m s^-1 × 2.00 ns x $\left| \frac{{{10}^{-9}}s}{1\,\,ns} \right|$ = 6.00 ×10^-1m = 0.600 m

Q.1.23 In a reaction A + B₂ → AB₂, Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 300 atoms of A + 200 molecules of B

(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv) 5 mol A + 2.5 mol B

(v) 2.5 mol A + 5 mol B

Ans: (i) 300 atoms of A + 200 molecules of B: According to the given reaction, 1 atom of A reacts with 1 molecule of B. therefore, 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.

(ii) 2 mol A + 3 mol B: According to the given reaction, 1 mol of A reacts with 1 mol of B. therefore, 2 mol of A will react with 2 mol of B. Hence, A is the limiting reactant

(iii) 100 atoms of A + 100 molecules of B: According to the given reaction, 1 atom of A combines with 1molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, there is no limiting reagent.

(iv) 5 mol A + 2.5 mol B:

According to the given reaction, 5 mol of A reacts with 2.5 mol of B. therefore, 2.5 mol of B will react with 2.5 mol of A. Hence, B is the limiting reactant

(v) 2.5 mol A + 5 mol B:

According to the given reaction, 2.5 mol of A reacts with 5 mol of B. therefore, 2.5 mol of A will react with 2.5 mol of B. Hence, A is the limiting reactant

Q.1.24 Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N_{2 (g)}+ H_{2 (g)}→ 2NH_{3 (g)}

(i) Calculate the mass of ammonia produced if 2.00 X 10³ g dinitrogen reacts with 1.00 X10³ g of dihydrogen.

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass?

Ans: (i) 1 mol of N₂(28 g) react with 3 mol (6 g) of H₂

So, 2000 g of N₂ will react with H₂ = $\frac{6\,g\,\,}{28\,g}$ x 2000 g = 428.6 g

But we are given amount of H₂ = 1000 g which is the excess reagent.

Thus, in this reaction N₂ will be completely consumed. Therefore, it is the limiting reagent.

Now,

2 mol of N₂ (28 g) produce NH₃ = 2 mol (34 g)

∴ 2000 g of N₂ will produce NH₃ = $\frac{34\,g\,\,}{28\,g}$x 2000 g = 2428.57 g

Hence, mass of NH₃ produced = 2428.57 g

(ii) H₂ will remain unreacted.

(iii) Mass left of H₂ unreacted = (Given mass – Mass used in reaction)

= (1000 – 428.6) = 571.4 g

Q.1.25 How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?

Ans: Molar mass of Na₂CO₃ = (2 × 23) + 12 + (3×16) = 106 g mol^–1

As we know that,

Mole (n) = $\frac{Given\,\,mass\,\,}{Molar\,\,mass}=\frac{w(g)\,}{M\left( \frac{g}{mol} \right)}$

Therefore, Mass (W) = Mole(n) X Molar mass (M)

(i) 0.50 mol Na₂CO₃ means, 0.50 mol of Na₂CO₃ contains (0.50 mol X 106 g mol^-1) 53 g of Na₂CO₃.

(ii) 0.50 M Na₂CO₃ means 0.50 mol, i.e., 53 g Na₂CO₃ are present in 1 litre of the solution.

Q.1.26 If 10 volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Ans: H₂ and O₂ react according to the equation

2H₂(g) + O₂(g) → 2H₂O(g)

Thus, 2 volumes of H₂ react with 1 volume of O₂ to produce 2 volumes of water vapour (2:1:2).

Hence, 10 volumes of H₂ will react completely with 5 volumes of O₂ to produce 10 volumes of water vapour.

Q.1.27 Convert the following into basic units:

(i) 28.7 pm

(ii) 15.15 pm

(iii) 25365 mg

Ans: (i) 28.7 pm

28.7 pm = 28.7 pm × $\left| \frac{{{10}^{-12}}m}{1\,\,pm} \right|$ = 2.87 × 10^-11 m

(ii) 15.15 pm

15.15 pm =15.15 pm × $\left| \frac{{{10}^{-12}}m}{1\,\,pm} \right|$ = 15.15 × 10^-12 m or 1.515 × 10^-11m

(iii) 25365 mg

25365 mg = 25365 mg × $\left| \frac{1\,\,g}{1000\,\,mg} \right|\times \left| \frac{1\,\,kg}{1000\,\,mg} \right|$ = 2.5365 × 10^-2 kg

Q.1.28 Which one of the following will have the largest number of atoms?

(i) 1 g Au (s)

(ii) 1 g Na (s)

(iii) 1 g Li (s)

(iv) 1 g of Cl₂(g)

(Atomic masses : Au = 197, Na = 23, Li = 7, Cl = 35 5 amu)

Ans: (i) 1 g Au (s

1 g Au = $\frac{1}{197\,}mol\,=\frac{1}{197\,}mol\,\times 6.023\times {{10}^{23}}\,\frac{atom}{mol}=3.06\times {{10}^{21}}\,atoms\,\,of\,\,Au$

(ii) 1 g Na (s)

1 g Na = $\frac{1}{23\,}mol\,=\frac{1}{23\,}mol\,\times 6.023\times {{10}^{23}}\,\frac{atom}{mol}$ = 26.2 x 10²¹ atoms of Na

(iii) 1 g Li (s)

1 g Li = $\frac{1}{7\,}mol\,=\frac{1}{7\,}mol\,\times 6.023\times {{10}^{23}}\,\frac{atom}{mol}$ = 86.0 x 10²¹ atoms of LI

(iv) 1 g of Cl₂(g)

1 g Cl₂ = $\frac{1}{71\,}mol\,=\frac{1}{71\,}mol\,\times 6.023\times {{10}^{23}}\,\frac{atom}{mol}$ = 8.48 x 10²¹ atoms of Cl₂

Thus 1 g of Li has the largest number of atoms.

Q.1.29 Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Ans: Mole fraction of C₂H₅OH= $\frac{Moles\,of\,{{C}_{2}}{{H}_{5}}OH}{Total\,\,moles\,}$

0.040 = $\frac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}\,}{{{n}_{{{C}_{2}}{{H}_{5}}OH}}+{{n}_{{{H}_{2}}O}}\,}$ eq. (i)

Number of moles in 1L of water, ${{n}_{{{H}_{2}}O}}=\frac{1000\,g}{18\,g\,mo{{l}^{-1}}\,}$= 55.55 mol

Substituting ${{n}_{{{H}_{2}}O}}$= 55.55 mol in eq.(i), we get

0.040 = $\frac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}\,}{{{n}_{{{C}_{2}}{{H}_{5}}OH}}+55.55\,}$

${{n}_{{{C}_{2}}{{H}_{5}}OH}}$ = 0.040 ( ${{n}_{{{C}_{2}}{{H}_{5}}OH}}+55.55$)

${{n}_{{{C}_{2}}{{H}_{5}}OH}}$ = (0.040 x ${{n}_{{{C}_{2}}{{H}_{5}}OH}}$) + (0.040 x 55.55)

${{n}_{{{C}_{2}}{{H}_{5}}OH}}$ = (0.040 x ${{n}_{{{C}_{2}}{{H}_{5}}OH}}$) + 2.222

${{n}_{{{C}_{2}}{{H}_{5}}OH}}$ – 0.040 ${{n}_{{{C}_{2}}{{H}_{5}}OH}}$ = 2.222

${{n}_{{{C}_{2}}{{H}_{5}}OH}}$ (1- 0.040) = 2.222

0.96 ${{n}_{{{C}_{2}}{{H}_{5}}OH}}$ = 2.222 mol

${{n}_{{{C}_{2}}{{H}_{5}}OH}}=\frac{2.222}{0.96}mol$

${{n}_{{{C}_{2}}{{H}_{5}}OH}}$= 2.314 mol

Hence, molarity of solution = $\frac{2.314\,mol}{1\,L}$ = 2.134 M or 2.134 Molar solution.

Q.1.30. What will be the mass of one 12C atom in g?

Ans. 1 mol of ¹²C atoms = 6.022 x 10²³ atoms = 12 g

Thus, 6.022 x 10²³ atoms of ¹²C have mass = 12 g

∴ 1 atom of ¹²C will have mass = $\frac{12\,g}{6.023\times {{10}^{23}}}$ = 1.9927×10^-23 g

Q.1.31 How many significant figures should be present in the answer of the following calculations?

(i) $\frac{0.02856\times 298.15\times 0.112}{0.5785}$

(ii) 5 X 5.364

(iii) 0.0125 + 0.7864 + 0.0215

Ans: (i) $\frac{0.02856\times 298.15\times 0.112}{0.5785}$

Least precise no. of calculation has three significant figures (in 0.112)

Hence, no. of significant figures = 3

(ii) 5 x 5.365

The second term has 4 significant figures. Hence, the answer should have 4 significant figures.

(iii) 0.012 + 0.7864 + 0.0215

As the least no. of decimal place in each term is 4. Hence, the answer should have 4 significant figures.

Q.1.32: Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Ans :

Q.1.33 Calculate the number of atoms in each of the following

(i) 52 moles of Ar

(ii) 52 u of He

(iii) 52 g of He.

Ans: (i) 52 moles of Ar

1 mole of Ar = 6.023 x 10²³ atoms

∴ 52 mol of Ar = 52 x 6.023×10²³ atoms = 3.131 x 10²⁵ atoms

(ii) 52 u of He

1 atom of He = 4 u of He or

∵ 4 u of He = 1 atom of He

∴ 1 u of He = $\frac{1}{4}$ atom of He

∴ 52 u of He = $\frac{1}{4}$x 52 = 13 atom of He

(iii) 52 g of He

∵4 g (1mol) of He = 6.023 x 10²³ atoms of He

∴ 52 g of He = $\frac{6.023\times {{10}^{23}}}{4}$x 52 = 7.8286 x10²⁴atoms of He

Q.1.34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate

(i) empirical formula,

(ii) molar mass of the gas,

(iii) molecular formula.

Ans: Amount of C in 3.38 g CO₂ = $\frac{12\,g}{44\,g}$x 3.38 g = 0.9218 g

Amount of H in 0.690 g H₂O = $\frac{2\,g}{18\,g}$x 0.690 g = 0.0767 g

As compound contains only C and H, therefore, total mass of

the compound = 0.9218 + 0.0767 g = 0.9985 g

% of C in compound = $\frac{0.9218\,g}{0.9985\,g}$x 100 = 92.32 %

% Of H in compound = $\frac{0.0767\,g}{0.9985\,g}$x 100 = 7.68 %

Calculation of Empirical Formula:

Empirical formula = CH

Empirical formula mass of CH = (12+1) = 13 g

Now,

10.0 L of the gas at STP weigh = 11.6 g

22.4 L of the gas at STP weigh = $\frac{11.6}{10.0}$x 22.4 = 25.98 g ≈ 26 g

Compound of Molar mass (Given) = 26 g mol^-1

n = $\frac{Molar\,\,mass}{Empirical\,\,Formula\,\,mass}$ = $\frac{26\,}{13}$ = 2

Hence, molecular formula = n x CH = 2 x CH = C₂H₂

Q.1.35 Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO_{3 (s)} + 2HCl _(aq) → CaCl_2(aq) + CO_2(g) + H₂O_(l)

What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?

Ans: 0.75 M of HCI means 0.75 mol of HCI are present in 1 L of solution.

1000 mL of 0.75 M HCl contain HCl = 0.75 mol

= 0.75 mol x 36.5 g/mol = 27.375 g

Thus, 1000 mL of solution contains 27.375 g of HCI

Therefore, amount of HCI present in 25 mL of solution

= $\frac{27.375\,g}{1000\,ml}$x 25 ml = 0.6844 g

According to given chemical reaction,

CaCO_{3 (s)} + 2HCl _(aq) → CaCl_2(aq) + CO_2(g) + H₂O_(l)

2 mol of HCI (2×36.5 =73 g) react with 1 mol of CaCO₃ = 100 g

Therefore, 0.6844 g HCl will react completely CaCO₃

= $\frac{100\,g}{73\,g}\times 0.6844$x 25 ml = 0.9375 g