Ncert Exercises & Solutions

2.13 The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas?

Molar mass of ethane (C₂H₆) = 2 × 12 + 6 × 1 = 30 g mol⁻¹

$∴$ Number of moles present in 6.56 x 10-2 g of ethane= $\frac{6.56 \times 10^{- 2}}{30} = 2.187 \times 10^{- 3} m o l$

According to Henry's law,

$p = K_{H} x$
$\Rightarrow 1 bar = K_{H} . \frac{2.187 \times 10^{- 3}}{2.187 \times 10^{- 3} + x}$
$\Rightarrow 1 bar = K_{H} \frac{2.187 \times 10^{- 3}}{x}$
$\Rightarrow K_{H} = \frac{x}{2.187 \times 10^{- 3}} bar (since x > 2.187 \times 10^{- 3})$

Number of moles present in $5.00 \times 10^{- 3}$ g of ethane= $\frac{5.00 \times 10^{- 2}}{30}$ mol

$= 1.67 \times 10^{- 3}$ mol

According to Henry's law.

$p = K_{H} x$
$= \frac{x}{2.187 \times 10^{- 3}} \times \frac{1.67 \times 10^{- 3}}{(1.67 \times 10^{- 3}) + x}$
$= \frac{x}{2.187 \times 10^{- 3}} \times \frac{1.67 \times 10^{- 3}}{x}$

=0.764

Hence, the partial pressure of the gas shall be 0.764 bar.

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