From the following data calculate the atomic weight of carbon. Compound Vapour density % of C CO 14 42 . 8 CS 2 38 15 . 8 C 2 H 2 14 85 . 7 C 3 H 4 22 81 . 4 C 6 H 6 39 92 . 3

Since we know that molecular weight Molecular mass = 2 × V . d where V . d = Vapour density If we consider W is the weight of carbon Then, (I) For CO , the number of carbon atom is equal to one and vapour density of CO is equal to 14 . Hence, molecular weight = 2 × V . d = 2 × 14 = 28 g . Since W is the atomic weight of carbon ∴ W = 28 × 42 . 8 % = 28 × 42 . 8 100 = 11 . 98 ≃ 12 g . Similarly, (II) In CS 2 , the number of carbon atom is equal to one and vapour density of CS 2 is equal to 38 . Hence, molecular weight = 2 × V . d = 2 × 38 = 76 g If W is the weight of carbon ∴ W = 76 × 15 . 8 % W = 76 × 15 . 8 100 = 12 W = 12 g . Similarly, (III) In C 2 H 2 , the number of carbon atom is equal to two and vapour density of C 2 H 2 is equal to 14 . Hence, molecular weight = 2 × V . d = 2 × 14 = 28 g . If W is the atomic weight of carbon Since two carbon atoms hence, 2 W = 28 × 85 . 7 % = 28 × 85 . 7 100 W = 11 . 998 W ≃ 12 g . Similarly, (IV) In C 3 H 4 , the number of carbon atom is equal to three and vapour density of C 3 H 4 is equal to 22 . Hence, molecular weight = 2 × V . d = 2 × 22 = 44 g . If W is the weight of carbon Since three carbon atoms hence, ∴ 3 W = 44 × 81 . 4 100 W = 11 . 94 g W ≃ 12 g Similarly, (V) In C 6 H 6 , the number of carbon atom is equal to six and vapour density of C 6 H 6 is equal to 39 . Hence, molecular weight = 2 × V . d = 2 × 39 = 78 g . If W is the weight of carbon Since six carbon atom in C 6 H 6 6 W = 78 × 92 . 3 % W = 78 × 92 . 3 6 × 100 W = 11 . 99 g W ≃ 12 g .

Ramendra C Mukerjee Solutions for Chapter: Atomic Weight, Exercise 1: PROBLEMS

Author:Ramendra C Mukerjee

Ramendra C Mukerjee Chemistry Solutions for Exercise - Ramendra C Mukerjee Solutions for Chapter: Atomic Weight, Exercise 1: PROBLEMS

Attempt the practice questions on Chapter 4: Atomic Weight, Exercise 1: PROBLEMS with hints and solutions to strengthen your understanding. Modern Approach to Chemical Calculations solutions are prepared by Experienced Embibe Experts.

Questions from Ramendra C Mukerjee Solutions for Chapter: Atomic Weight, Exercise 1: PROBLEMS with Hints & Solutions

EASY

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 4 - Atomic Weight>PROBLEMS>Q 4

Naturally occurring boron consists of two isotopes whose atomic weights are $10.01$ and $11.01 .$ The atomic weight of natural boron is $10.81 .$ Calculate the percentage of each isotope in natural boron.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 4 - Atomic Weight>PROBLEMS>Q 5

In an experiment, pure carbon monoxide was passed over red-hot copper oxide. ${CO}_{2},$ so produced, weighed $0.88 g$ and the weight of copper oxide was reduced by $0.3232 g$ . Calculate the atomic weight of carbon(the nearest integer in amu) [molar mass of copper $= 63.5 g$ ]. Give answer to the nearest integer value.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 4 - Atomic Weight>PROBLEMS>Q 7

From the following data calculate the atomic weight of carbon.

$\begin{matrix} Compound & Vapour density & % of C \\ CO & 14 & 42.8 \\ {CS}_{2} & 38 & 15.8 \\ C_{2} H_{2} & 14 & 85.7 \\ C_{3} H_{4} & 22 & 81.4 \\ C_{6} H_{6} & 39 & 92.3 \end{matrix}$

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 4 - Atomic Weight>PROBLEMS>Q 11

$0.096 g$ of stannic chloride gave $25.84 mL$ of its vapour at $120 ° C$ and $350 mm$ pressure. If the chloride contains $54.6 %$ of chlorine and tin has the valency equal to $4,$ what will be the atomic weight of tin?

Round-off the answer to the nearest integer.

MEDIUM

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 4 - Atomic Weight>PROBLEMS>Q 13

${Cu}_{2} S$ and ${Ag}_{2} S$ are isomorphous. The percentages of sulphur in these compounds are $20.14 %$ and $12.94 %$ respectively. If the atomic weight of $Cu$ is $63.5,$ calculate the atomic weight of $Ag$ .

MEDIUM

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 4 - Atomic Weight>PROBLEMS>Q 14

The natural titanium oxide, known as rutile and containing $39.95 %$ of oxygen, is isomorphous with ${SnO}_{2},$ known as cassiterite. Calculate the atomic weight of titanium.

Give the answer up to first place of decimal.

MEDIUM

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 4 - Atomic Weight>PROBLEMS>Q 15

The atomic weight of sulphur was determined by decomposing $6.2984 g$ of ${Na}_{2} {CO}_{3}$ with $H_{2} {SO}_{4}$ and weighing the resultant ${Na}_{2} {SO}_{4}$ formed. The weight was found to be $8.43 g .$ Taking the atomic weights of $C, O$ and $Na$ as $12.01, 15.99$ and $22.99$ respectively, calculate the atomic weight of sulphur.

Give the answer up to two places of decimal.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 4 - Atomic Weight>PROBLEMS>Q 17

In a chemical determination of the atomic weight of vanadium, $2.8934 g$ of pure ${VOCl}_{3}$ was allowed to undergo a set of reactions as a result of which all the chlorine contained in this compound was converted to $AgCl$ which weighed $7.18 g$ . Calculate the atomic weight of vanadium. $(Ag = 107.868, Cl = 35.45, O = 16)$

Ramendra C Mukerjee Solutions for Chapter: Atomic Weight, Exercise 1: PROBLEMS

Ramendra C Mukerjee Chemistry Solutions for Exercise - Ramendra C Mukerjee Solutions for Chapter: Atomic Weight, Exercise 1: PROBLEMS

Questions from Ramendra C Mukerjee Solutions for Chapter: Atomic Weight, Exercise 1: PROBLEMS with Hints & Solutions

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