**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

Toggle**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) H _{2}O **

**(ii) CO _{2} **

**(iii) CH _{4}**

**Ans:** (i) Molecular mass of H_{2}O = 2 (1.008 amu) + 16.00 amu = 18 016 amu

(ii) Molecular mass of CO_{2} = 12.01 amu + 2 (16.00 amu) = 44.01 amu

(iii) Molecular mass of CH_{4} = 12.01 amu + 4 (1.008 amu) = 16 .042

**Q.****1.2: Calculate the mass per cent of different elements present in sodium sulphate (Na _{2}SO_{4}).**

**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 Na_{2}SO_{4} = (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 = Fe_{2}O_{3}.

**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 CO_{2 }is produced.

(ii) As only 16 g of dioxygen is available, it can combine only with half mole (0.5 mol) of carbon. Hence, CO_{2} 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, CO_{2} produced = ½ mol or 22 g. here , O_{2 }is the limiting agent.

**Q.****1.5: Calculate the mass of sodium acetate (CH _{3}COONa) 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 (HNO_{3}) = 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 HNO_{3}, 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 (CuSO _{4})?**

**Ans: **Given, Copper sulphate (CuSO_{4}) = 100 g

Atomic mass of Cu = 63.5 amu

1 mole of CuSO_{4} contains = 1 mole of Cu

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

∵ Amount of Cu can be obtained from 159.5 g CuSO_{4} = 63.5 g

∴ Amount of Cu can be obtained from 100 g CuSO_{4} = $\frac{63.5}{159.5}$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 Fe_{2}O_{3}

The empirical formula can be calculated as :

Therefore, Empirical formula = Fe_{2}O_{3}.

Now, calculation of Empirical formula mass of Fe_{2}O_{3}

= (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 Fe_{2}O_{3}

Hence, Molecular formula = Fe_{2}O_{3} (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 (C _{2}H_{6} ), 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 C_{2}H_{6} contains moles of C atoms = 2 mole

∴ 3 moles of C_{2}H_{6} contains moles of C atoms = 3 x 2 = 6 mole of C atoms

(ii) ∵ 1 mole of C_{2}H_{6} contains moles of H atoms = 6 mole

∴ 3 moles of C_{2}H_{6} contains moles of H atoms = 3 x 6 = 18 mole of H atoms

(iii) ∵ 1 mole of C_{2}H_{6} contains number of molecules of ethane = 6.023 × 10^{23}

∴ 3 moles of C_{2}H_{6} contains number of molecules of ethane = 3 X 6.023 × 10^{23} = 18.069 X10^{23 }

**Q.****1.11 What is the concentration of sugar (C _{12}H_{22}O_{11}) 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 (CH_{3}OH) = 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 M_{1} x V_{1 }= M_{2} x V_{2}

24.78 x V_{1} = 0.25 x 2.5 L

V_{1} = 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 = 10^{6}, giga = 10^{9}, 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, CHCl _{3}, 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 x10^{5}

(iii) 8.008 x 10^{3}

(iv) 5.000 x 10^{2}

(v) 6.0012 x 10^{0}

**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 (N_{2}) 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 10 ^{8} 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 10^{8} m s^{–1} ) X 2.00 ns or

= 3.0 ×10^{8 }m s^{-1} × 2.00 ns x $\left| \frac{{{10}^{-9}}s}{1\,\,ns} \right|$ = 6.00 ×10^{-1 }m = 0.600 m

**Q.****1.23 In a reaction A + B _{2} **

**→**

**AB**

_{2}**, 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 10 ^{3} g dinitrogen reacts with 1.00 X10^{3} 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 N _{2}(28 g) react with 3 mol (6 g) of H_{2} **

So, 2000 g of N_{2} will react with H_{2} = $\frac{6\,g\,\,}{28\,g}$ x 2000 g = 428.6 g

But we are given amount of H_{2} = 1000 g which is the excess reagent.

Thus, in this reaction N_{2} will be completely consumed. Therefore, it is the limiting reagent.

Now,

2 mol of N_{2} (28 g) produce NH_{3} = 2 mol (34 g)

∴ 2000 g of N_{2} will produce NH_{3} = $\frac{34\,g\,\,}{28\,g}$x 2000 g = 2428.57 g

Hence, mass of NH_{3} produced = 2428.57 g

**(ii) H _{2} will remain unreacted. **

**(iii) Mass left of H _{2} unreacted** = (Given mass – Mass used in reaction)

= (1000 – 428.6) = 571.4 g

**Q.****1.25 How are 0.50 mol Na _{2}CO_{3} and 0.50 M Na_{2}CO_{3} different?**

**Ans: **Molar mass of Na_{2}CO_{3} = (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 Na_{2}CO_{3} means, 0.50 mol of Na_{2}CO_{3} contains (0.50 mol X 106 g mol^{-1}) 53 g of Na_{2}CO_{3}.

(ii) 0.50 M Na_{2}CO_{3} means 0.50 mol, i.e., 53 g Na_{2}CO_{3} 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 vol****umes of water vapour would be produced?**

**Ans:** H_{2} and O_{2} react according to the equation

2H_{2}(g) + O_{2 }(g) → 2H_{2}O(g)

Thus, 2 volumes of H_{2} react with 1 volume of O_{2} to produce 2 volumes of water vapour (2:1:2).

Hence, 10 volumes of H_{2} will react completely with 5 volumes of O_{2} 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^{-11 }m

**(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 Cl _{2}(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 10^{21} 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 10^{21} atoms of LI

**(iv) 1 g of Cl _{2}(g)**

1 g Cl_{2} = $\frac{1}{71\,}mol\,=\frac{1}{71\,}mol\,\times 6.023\times {{10}^{23}}\,\frac{atom}{mol}$ = 8.48 x 10^{21} atoms of Cl_{2}

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 C_{2}H_{5}OH= $\frac{Moles\,of\,{{C}_{2}}{{H}_{5}}OH}{Total\,\,moles\,}$

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 ^{12}C atoms = 6.022 x 10^{23} atoms = 12 g

Thus, 6.022 x 10^{23} atoms of ^{12}C have mass = 12 g

∴ 1 atom of ^{12}C 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 10^{23} atoms

∴ 52 mol of Ar = 52 x 6.023×10^{23} atoms = 3.131 x 10^{25} 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 10^{23} atoms of He

∴ 52 g of He = $\frac{6.023\times {{10}^{23}}}{4}$x 52 = 7.8286 x10^{24 }atoms 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 CO_{2} = $\frac{12\,g}{44\,g}$x 3.38 g = 0.9218 g

Amount of H in 0.690 g H_{2}O = $\frac{2\,g}{18\,g}$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 = C_{2}H_{2}

**Q.****1.35 Calcium carbonate reacts with ****aqueous HCl to give CaCl _{2} and CO_{2} according to the reaction, CaCO_{3 (s)} + 2HCl _{(aq)} **

**→**

**CaCl**

_{2(aq)}+ CO_{2(g)}+ H_{2}O_{(l)}**What mass of CaCO _{3} 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)} + H_{2}O_{(l)}

2 mol of HCI (2×36.5 =73 g) react with 1 mol of CaCO_{3} = 100 g

Therefore, 0.6844 g HCl will react completely CaCO_{3}

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

**Q.****1.36 Chlorine is prepared in the laboratory by treating ****manganese dioxide (MnO _{2}) with aqueous hydrochloric acid according to the reaction: **

**4 HCl _{(aq)} + MnO_{2(s)} **

**→**

**2H**

_{2}O_{(l)}+ MnCl_{2(aq)}+ Cl_{2(g)}

**How many grams of HCl react with 5.0 g of manganese dioxide?**

** ****Ans: **1 mol of MnO_{2} = 55 + (2×16) = 87 g

4 mol of HCI = 4 x 36.5 = 146 g

∵ 1 mol (87 g) of MnO_{2} reacts with 4 mol of HCI (4x 36.5 gm) = 146 g

∴ 5 g of MnO_{2} will react with = $\frac{146}{87}$x 5 = 8.40 g HCl

Hence, 8.4 g of HCI will react with 5 g of MnO_{2}.