$(a)\;1\;atm\qquad(b)\;0.28\;atm\qquad(c)\;0.56\;atm\qquad(d)\;1.2\;atm$

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Partial pressure of $O_2$ , $P'_{O_2} = P_m\times mole\; fraction$

Since $P_m$ = 1 atm (given NTP condition)

$P'_{O_2} = 1\times mole\;fraction$-----(i)

For mixture

$PV = \large\frac{w}{m}RT$

$m = \large\frac{w}{VP}RT$

$m=\large\frac{1.3\times0.0821\times273}{1}(\therefore \large\frac{w}{V}=1.3g/l)$

$\therefore \text{mol.wt of mixture} = 29.137$

$If \;n_1 \;and\; n_2\; are\; mole\; of\; O_2\; and\; N_2\; respectively$

$Now , \large\frac{32\times n_1+28\times n_2}{(n_1+n_2)}=29.137$

$\large\frac{28n_1+28n_2}{n_1+n_2} + \large\frac{4n_1}{n_1+n_2} = 29.137$

$\therefore \large\frac{n_1}{n_1+n_2} = \large\frac{29.137-28}{4} = 0.28$

$\therefore Mole\;fraction\;of\;O_2 = 0.28$

$\therefore By\; Eq (1) , P'_{O_2} = 0.28 \;atm$

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