A vessel contains a mixture consisting of m $_{1}$ - 7 g of nitrogen (M $_{1}$ = 28) and m $_{2}$ = 11 g of carbon dioxide (M $_{2}$ = 44) at temperature T - 300 K and pressure P $_{0}$ = 1 atm. The density of the mixture is

1.46 g p e r l i t r e

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2.567 g p e r l i t r e

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3.752 g p e r l i t r e

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4.572 g p e r l i t r e

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Solution

The correct option is A $1.46 g p e r l i t r e$
Let the volume occupied $= V$
By Dalton's law of partial pressure
$\frac{P_{n i t}}{P_{0}} = \frac{n_{n i t}}{n_{n i t} + n_{c a r b}}$
No. of moles of Nitrogen $η_{n i t} = \frac{M_{1}}{M_{n i t}} = \frac{7}{28} = 0.25 m o l$
No. of moles of carbon $η_{c a r b} = \frac{M_{2}}{M_{c a r b}} = \frac{11}{4} = 0.25 m o l$
Thus,
$P_{n i t} = P_{0} \times \frac{0.25}{0.25 \times 0.25} = P_{0} / 2 = 0.5 a t m$
From ideal gas equation
$V = \frac{n R T}{P} = \frac{0.25 \times 8.314 \times 290}{0.5 \times 101325} = 0.0119 m o l e$
Total mixture $m = (7 + 11) \times 10^{- 11} k g$
Thus density s $P = \frac{m}{V} = \frac{(7 + 11) \times 10^{- 3}}{0.0119} \approx 1.46 k g / m^{3}$
Option A is correct.

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Similar questions

m_{1} = 7 k g

(M_{1} = 28)

m_{2} = 11 g

(M_{2} = 44)

T = 300 K

P_{0} = 1 a t m

(= 1.01 \times 10^{5} N / m^{2})

(R = 8.31 J / m o l - K)

m_{1} =

m_{2} =

=

p_{0} =

m^{3}

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