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JEE MAIN 2022
29-07-2022 S2
Question
' x ' g of molecular oxygen $\left(\mathrm{O}_2\right)$ is mixed with 200 g of neon $(\mathrm{Ne})$. The total pressure of the non-reactive mixture of $\mathrm{O}_2$ and Ne in the cylinder is 25 bar. The partial pressure of Ne is 20 bar at the same temperature and volume. The value of ' $x$ ' is $\_\_\_\_$ . [Given: Molar mass of $\mathrm{O}_2=32 \mathrm{~g} \mathrm{~mol}^{-1}$. Molar mass of $\mathrm{Ne}=20 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
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\begin{aligned} & \mathrm{O}_2+\mathrm{Ne} \\ & \mathrm{X}_{\mathrm{gm}} 200 \mathrm{gm} \\ & \mathrm{P}_{\text {total }}=25 \mathrm{bar} ; \mathrm{P}_{\mathrm{Ne}}=20 \\ & \mathrm{P}_{\mathrm{O}_2}+\mathrm{P}_{\mathrm{Ne}}=25 \\ & \mathrm{P}_{\mathrm{O} 2}=25-20=5 \mathrm{bar} \\ & \quad 5=\frac{\frac{\mathrm{x}}{32}}{\frac{\mathrm{x}}{32}+\frac{200}{20}} \times 25 \\ & \quad \frac{1}{5}=\frac{\frac{\mathrm{x}}{32}}{\frac{\mathrm{x}}{32}+10} & \frac{1}{5}=\frac{x \times 32}{32(x+320)} \\ & 5 x=x+320 \\ & 4 x=320 \\ & x=\frac{320}{4}=80 \mathrm{gm} \end{aligned}
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