Ncert Exercises & Solutions

Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 10⁵ mm Hg. The solubility of methane in benzene at 298 K under 760 mm Hg is -

1. 1.78 x 10^-3

2. 1.78 x 10^-5

3. 2.98 x 10^-4

4. 3.78 x 10^-3

Hint: Formula of Henry's law; P=

k_{H} X_{solute}

Step 1:

The given values are as follows:

Here, p = 760 mm Hg

k_H = 4.27 × 10⁵mm Hg

According to Henry’s law,

$p = k_{H} x$

$Step 2 :$

$Calculate the mole fraction of solute as follows :$
$\Rightarrow x = \frac{p}{k_{H}}$
$= \frac{760 mm Hg}{4.27 \times 10^{5} mm Hg}$
$= 177.99 \times 10^{- 5}$
$= 1.78 \times 10^{- 3} (approximately)$

Hence, the mole fraction of methane in benzene is 1.78 × 10⁻³.

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