SATHEE: Some Basic Concepts Of Chemistry Question 132

Some Basic Concepts Of Chemistry Question 132

Question: 23.2 g of an organic compound having molecular formula $C_{n} H_{2 n + 2}$ is burnt in excess of $O_{2} (g)$ initially taken in a 44.82 L steel vessel. Before reaction the gaseous mixture w at 273 K with pressure of 2 aim. After complete combustion and loss of considerable amount of heat the mixture of product and excess of had a temperature of 546 K and 4.6 atm pressure. The formula of compound is

Options:

A) $C_{5} H_{12}$

B) $C_{6} H_{10}$

C) $C_{3} H_{8}$

D) $C_{4} H_{10}$

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Idea This problem can be solved by using concept of ideal gas equation and chemical equation involved in combustion reaction. Students are advised to follow steps. Write chemical equation involved in combustion of hydrocarbon. Assume initial pressure = p Calculate increased pressure Then, calculate number of mole using ideal gas equation. Equation of combustion $C_{n} H_{2} n_{+ 2} + (\frac{3 n + 1}{2}) O_{2} \overset{}{\to} n C O_{2} + (n + 1) H_{2} O$ Initial pressure of $C_{n} H_{2} n_{+ 2}$ is p (assumed) Increase in pressure $= p [(2 n + 1) - 1 - (\frac{3 n + 1}{2})] = (\frac{n - 1}{2}) p$

$∵$ Mass of organic compound $= 14 \times n + 2$ 546 K and 4.6 atm or 273 K 2.3 atm Increase
$\Rightarrow$ $2.3 - 2 = 0.3$ atm $p = \frac{n R T}{V} = \frac{23.2}{M} \times (\frac{0.0821 \times 273}{44.82})$

$= \frac{23.2}{M} \times \frac{0.5}{44.82}$

$= \frac{116}{(14 n + 2)} \times \frac{(n - 1)}{2}$ also $= \frac{(n - 1)}{2} \times \frac{11.6}{14 n + 2} = 0.3$ On solving, $n = 4$ So, compound will be $C_{4} H_{4 \times 2 + 2} \Rightarrow C_{4} H_{10}$

Some Basic Concepts Of Chemistry Question 132

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Some Basic Concepts Of Chemistry Question 132

Options:

Answer:

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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