Question

At $27^{\circ} C$ in a $10\, L$ flask $4.0\, g$ of an ideal gaseous mixture containing. He (molar mass $4.0\, g\, mol ^{-1}$ ) and Ne (molar mass $20\, g\, mol ^{-1}$ ) has a pressure of $1.23$ atm. What is the mass $\%$ of neon $?\left(R=0.082\, L\, atm\, K ^{-1}\, mol ^{-1}\right)$

• A. 25.2
• B. 62.5
• C. 84.2
• D. 74.2

Answer 1

Answer: (B)

Given $T =27+273=300\, K$

$V =10\, L$

$W =W_{1}+W_{2}=4\, g$

$M_{( He )} =4$

$M_{( Ne )} =20$

$p =1.23\, atm$

$R =0.082\, L\, atm\, K ^{-1}\, mol ^{-1}$

(i) $\because pV =\left(n_{1}+ n _{2}\right) RT$

$=\left[\frac{W_{1}}{4}+\frac{W}{20}\right] R T$

$\therefore \left[\frac{W_{1}}{4}+\frac{W_{2}}{20}\right]=\frac{p V}{R T}$

$=\frac{1.23 \times 10}{0.082 \times 300}=0.5$

$25 W_{1}+5 W _{2}=50$

Or, $5 W_{1}+W_{2}=10$ ...(i)

Also (ii) $W_{1}+W_{2}=4$ ...(ii)

On subtracting Eq. (ii) from Eq. (i)

$4 W_{1}=6 \Rightarrow W_{1}=1.5$

and $W_{2}=4-1.5=2.5$

Hence, Mass $\%$ of $Ne$ (neon)

$=\frac{2.5 \times 100}{4}=62.5 \%$