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Chemistry Part-I

Some Basic Concepts of Chemistry

A welding fuel gas contains carbon and hydrogen only. Burning

Chemistry Part-I

Class 11

NCERT

Chapter 1: Some Basic Concepts of Chemistry

44 questions

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.

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 and (iii) molecular formula.

Calcium carbonate reacts with aqueous

H Cl

to give

C a Cl_{2}

and

CO_{2}

according to the reaction,

C a CO_{3} (s) + 2 H Cl (a q) \to C a Cl_{2} (a q) + CO_{2} (g) + H_{2} O (l)

What mass of

C a CO_{3}

is required to react completely with 25 mL of 0.75 M

H Cl

View all question

Chapter 2: Structure of Atom

85 questions

Chapter 3: Classification of Elements and Periodicity in Properties

50 questions

Chapter 4: Chemical Bonding and Molecular Structure

44 questions

Chapter 5: States of Matter

28 questions

Chapter 6: Thermodynamics and Thermochemistry

36 questions

Chapter 7: Equilibrium

101 questions

Question

Hard

Solving time: 8 mins

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 and (iii) molecular formula.

Views: 6,910 students

Updated on: Oct 5, 2023

Book:

Chemistry Part-I (NCERT)

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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 and (iii) molecular formula.

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is

\frac{12}{44} \times 3.38 = 0.92

g.
Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is

\frac{2}{18} \times 0.690 = 0.077

g.
The percentage of C is

\frac{0.92}{0.92 + 0.077} \times 100 = 92.3

%.
The percentage of H is

\frac{0.077}{0.92 + 0.077} \times 100 = 7.7

(i) The number of moles of carbon

= \frac{92.2}{12} = 7.7

.
The number of moles of hydrogen

= \frac{7.7}{1} = 7.7

.
The mole ratio

C : H = \frac{7.7}{7.7} = 1 : 1

.
Hence, the empirical formula of the welding fuel gas is CH.

(ii) The weight of 10.0 L of gas at N.T.P is 11.6 g.
22.4 L pf gas at N.T. P will weigh

\frac{11.6}{10.0} \times 22.4 = 26 g / m o l

This is the molar mass.

(iii) Empirical formula mass is

12 + 1 = 13

.
Molecular mass is 26.
The ratio, n of the molecular mass to empirical formula mass is

\frac{26}{13} = 2

.
The molecular formula is

2 (C H) = C_{2} H_{2}

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Practice more questions from Chemistry Part-I (NCERT)

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.

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 and (iii) molecular formula.

Calcium carbonate reacts with aqueous

H Cl

to give

C a Cl_{2}

and

CO_{2}

according to the reaction,

C a CO_{3} (s) + 2 H Cl (a q) \to C a Cl_{2} (a q) + CO_{2} (g) + H_{2} O (l)

What mass of

C a CO_{3}

is required to react completely with 25 mL of 0.75 M

H Cl

View all

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Question Text	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 and (iii) molecular formula.
Updated On	Oct 5, 2023
Topic	Some Basic Concepts of Chemistry
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 17
Upvotes	1703
Avg. Video Duration	8 min

Text solutionVerified

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