Q.34. A mixture of carbon dioxide and oxygen has volume , temperature , pressure and mass . The number of moles of carbon dioxide and oxygen gases in the mixture respectively are
(Assume both carbond

Question

Q.34. A mixture of carbon dioxide and oxygen has volume , temperature , pressure  and mass . The number of moles of carbon dioxide and oxygen gases in the mixture respectively are
(Assume both carbondioxide and oxygen gases behave like ideal gases) [R = 8.31J/mol.K]

FirstInTest · Accepted Answer

Correct answer: C. Given: $V = 8310 \mathrm{~cm}^3 = 8.31 	imes 10^{-3} \mathrm{~m}^3$, $T = 300 \mathrm{~K}$, $P = 100 \mathrm{~kPa} = 10^5 \mathrm{~Pa}$, $R = 8.31 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}$.
Using Ideal Gas Law: $PV = n_{	ext{total}}RT$.
$n_{	ext{total}} = \frac{PV}{RT} = \frac{10^5 	imes 8.31 	imes 10^{-3}}{8.31 	imes 300} = \frac{100}{300} = 0.333 \mathrm{~mol}$.
Let $n_1$ be moles of $\mathrm{CO}_2$ and $n_2$ be moles of $\mathrm{O}_2$.
$n_1 + n_2 = 0.333$ (Equation 1)
Molar mass of $\mathrm{CO}_2 = 12 + 2 	imes 16 = 44 \mathrm{~g/mol}$.
Molar mass of $\mathrm{O}_2 = 2 	imes 16 = 32 \mathrm{~g/mol}$.
Total mass $m = n_1 M_1 + n_2 M_2 = 44n_1 + 32n_2 = 13.2 \mathrm{~g}$ (Equation 2)
From (1), $n_2 = 0.333 - n_1$.
Substitute into (2): $44n_1 + 32(0.333 - n_1) = 13.2$
$44n_1 + 10.656 - 32n_1 = 13.2$
$12n_1 = 13.2 - 10.656 = 2.544$
$n_1 = \frac{2.544}{12} \approx 0.212 \mathrm{~mol}$.
$n_2 = 0.333 - 0.212 = 0.121 \mathrm{~mol}$.
So, $n_1 \approx 0.21 \mathrm{~mol}$ and $n_2 \approx 0.12 \mathrm{~mol}$. The closest option is (3) 0.21 and 0.12.

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