A vessel contains a mixture consisting of $m_{1} = 7 k g$ of nitrogen $(M_{1} = 28)$ and $m_{2} = 11 g$ of carbon dioixide $(M_{2} = 44)$ at temeprature $T = 300 K$ and pressure $P_{0} = 1 a t m$ . The density of the mixture is:

1.446 g

per litres

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2.567 g

per litre

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3.752 g

per litre

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4.572 g

per litre

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Solution

The correct option is A $1.446 g$ per litres
Let V is the volume of the vessel.
Now, let $p_{1}$ and $p_{2}$ be the partial pressure, then using gas law:
$p_{1} V = \frac{m_{1}}{M_{1}} R T$
$p_{2} V = \frac{m_{2}}{M_{2}} R T, p_{0} = p_{1} + p_{2}$
$p_{0} = (\frac{m_{1}}{M_{1}} + \frac{m_{2}}{M_{2}}) \frac{R T}{V}$
$V = (\frac{m_{1}}{M_{1}} + \frac{m_{2}}{M_{2}}) \frac{R T}{p_{0}}$
$∵ ρ_{m i x} = \frac{(m_{1} + m_{2})}{V}$
$ρ_{m i x} = \frac{(m_{1} + m_{2}) M_{1} M_{2}}{(m_{1} M_{2} + m_{2} M_{1})} \times \frac{p_{0}}{R T}$
Substituting values,
$ρ_{m i x} = \frac{(7 + 11) \times 28 \times 44 \times 10^{- 3}}{(7 \times 44 + 11 \times 28))} \times \frac{10^{5}}{8.3 \times 300}$
$= 1.446 p e r l i t r e$
Option A is correct.

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