At the sea level, the dry air mass percentage composition is given as nitrogen gas : $70.0$, oxygen gas : $27.0$ and argon gas : $3.0$. If total pressure is $1.15$ atm, then calculate the ratio of followings respectively :
(i) partial pressure of nitrogen gas to partial pressure of oxygen gas
(ii) partial pressure of oxygen gas to partial pressure of argon gas
(Given : Molar mass of N, O and Ar are $14,16,$ and $40\mathrm{g}{\mathrm{mol}}^{-1}$ respectively)

[JEE Main 2025, 7 Apr (Shift 1)]

Question

At the sea level, the dry air mass percentage composition is given as nitrogen gas : $70.0$, oxygen gas : $27.0$ and argon gas : $3.0$. If total pressure is $1.15$ atm, then calculate the ratio of followings respectively :
(i) partial pressure of nitrogen gas to partial pressure of oxygen gas
(ii) partial pressure of oxygen gas to partial pressure of argon gas
(Given : Molar mass of N, O and Ar are $14,16,$ and $40\mathrm{g}{\mathrm{mol}}^{-1}$ respectively)

[JEE Main 2025, 7 Apr (Shift 1)]

YESPYQ · Accepted Answer

Correct answer: d) 2.96, 11.2. $\frac{{P}_{{N}_{2}}}{{P}_{{O}_{2}}}=\frac{{X}_{{N}_{2}}.{P}_{T}}{{X}_{{O}_{2}}.{P}_{T}}=\frac{n{N}_{2}}{{n}_{{O}_{2}}}$$=\frac{70/28}{27/32}=2.96$$\frac{{P}_{{O}_{2}}}{{P}_{Ar}}=\frac{{n}_{{O}_{2}}}{{n}_{Ar}}=\frac{27/32}{3/40}=11.25$

At the sea level, the dry air mass percentage composition is given as nitrogen gas : \(70.0\), oxygen gas : \(27.0\) and…

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