Question

A closed vessel contains a mixture of two gases Neon & Argon the total mass of mixture is 28 gm.The partial pressure due to Argon and neon are 4atm and 12atm respectively.The mass of individual gases in vessel is$(M_{neon}=20,M_{argon}=40,R=8.3J/mol-k)$

• A. $4gm,24gm$
• B. $1gm,27gm$
• C. $6gm,22gm$
• D. $2gm,26gm$

Answer 1

The correct option is A $4gm,24gm$

Total Pressure = $P_{Argon}+P_{Neon}=16atm$

Let $x$ number of gas molecules be responsible for $16atm$ pressure

$\therefore$ Number of Argon's gas molecules $=\frac{4}{16}\times x=\frac{x}{4}$

Number of Neon's gas molecules $=\frac{12}{16}\times x=\frac{3x}{4}$

Mass of one molecule of Argon $=\frac{20g}{6.02\times {10}^{23}}$

$\therefore$Mass of $\frac{x}{4}$ molecules $=\frac{x}{4}\times \frac{20}{6.02\times {10}^{23}}$

Mass of one molecule of Neon $=\frac{40}{6.02\times {10}^{23}}$

$\therefore$Mass of $\frac{3x}{4}$ molecules $=\frac{3x}{4}\times \frac{40}{6.02\times {10}^{23}}$

Given,

$\frac{x}{4}\times \frac{20}{6.02\times {10}^{23}}+$ $\frac{3x}{4}\times \frac{40}{6.02\times {10}^{23}}=28$

We calculate, $x=\frac{2\times 4\times 6.02\times {10}^{23}}{10}$

$\frac{x}{4}$ of Argon weighs $4g$

$\frac{3x}{4}$ of Neon weighs $24g$