SATHEE: Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry - Result Question 57

####56. (a) $1.0 L$ of a mixture of $C O$ and $C O_{2}$ is taken. This mixture is passed through a tube containing red hot charcoal. The volume now becomes $1.6 L$ . The volumes are measured under the same conditions. Find the composition of mixture by volume.

(b) A compound contains 28 per cent of nitrogen and 72 per cent of a metal by weight. 3 atoms of metal combine with 2 atoms of nitrogen. Find the atomic weight of metal.

$(1980, 5 M)$

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Solution:

(a) After passing through red-hot charcoal, following reaction occurs

$C (s) + C O_{2} (g) ⟶ 2 C O (g)$

If the $1.0 L$ original mixture contain $x$ litre of $C O_{2}$ , after passing from tube containing red-hot charcoal, the new volumes would be :

$2 x$ (volume of $C O$ obtained from $C O_{2}) + 1$

$\begin{aligned} \Rightarrow x (original CO) & = 1 + x = 1.6 (given) x & = 0.6 \end{aligned}$

Hence, original $1.0 L$ mixture has $0.4 L C O$ and $0.6 L^{-}$ of $C O_{2}$ , i.e. $40$ and $60$ by volume.

(b) According to the given information, molecular formula of the compound is $M_{3} N_{2}$ . Also, 1.0 mole of compound has $28 g$ of nitrogen. If $X$ is the molar mass of compound, then :

$\begin{matrix} X \times \frac{28}{100} = 28 \Rightarrow X = 100 = 3 \times Atomic weight of M + 28 \Rightarrow Atomic weight of M = \frac{72}{3} = 24 \end{matrix}$

Some Basic Concepts of Chemistry - Result Question 57

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Some Basic Concepts of Chemistry - Result Question 57

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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