At $27^{\circ} \mathrm{C}$ in a $10 \mathrm{~L}$ flask $4.0 \mathrm{~g}$ of an ideal gaseous mixture…

At $27^{\circ} \mathrm{C}$ in a $10 \mathrm{~L}$ flask $4.0 \mathrm{~g}$ of an ideal gaseous mixture containing. He (molar mass $4.0 \mathrm{~g} \mathrm{~mol}^{-1}$ ) and $\mathrm{Ne}$ (molar mass $\left.20 \mathrm{~g} \mathrm{~mol}^{-1}\right)$ has a pres sure of $1.23 \mathrm{~atm}$. What is the mass $\%$ of neon ? $\left(R=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
  1. 25.2
  2. 62.5
  3. 84.2
  4. 74.2

Solution

Given $ \begin{aligned} & T=27+273=300 K \\ & V=10 L \end{aligned} $
$\begin{aligned} & \text { Molar mass }(\mathrm{He})=4 \\ & \text { Molar mass }(\mathrm{Ne})=20 \\ & P=123 \mathrm{~atm} \\ & R=0.082 \mathrm{~L} \text { atm K} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\end{aligned}$ $\begin{aligned} & \text { (i) } \quad \because \quad P V=\left(n_1+n_2\right) R T=\left[\frac{W_1}{4}+\frac{W_2}{20}\right] R T \\ & \therefore \quad\left[\frac{W_1}{4}+\frac{W_2}{20}\right]=\frac{1.23 \times 10}{0.082 \times 300}=0.5 \\ & 25 W_1+5 W_2=50\end{aligned}$
On solving equation (ii) with the help of eq. $(i)$ $ \begin{aligned} & 5 W_1+4-W_1=10 \Rightarrow \quad W_1=1.5 \\ & W_2=2.5 \\ & \text { Mass \% of neon }=\frac{2.5 \times 100}{4}=62.5 \% \end{aligned} $

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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