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Q.

When a gas is bubbled through water at 298 K, a very dilute solution of the gas is obtained. Henry’s law constant for the gas at 298 K is 100 kbar. If the gas exerts a partial pressure of 1 bar, the number of millimoles of the gas dissolved in one litre of water is

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a

0.0555

b

5.55

c

0.555

d

55.5

answer is A.

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Detailed Solution

$K_{H} = 100 kbar = 10^{5} bar, P = 1 bar$
$\begin{array}{l} P = K_{H} \times x_{A} \\ x_{A} = \frac{P}{K_{H}} = \frac{1}{100 \times 10^{3}} = 10^{- 5} \end{array}$
Moles of water= $\frac{1000}{18}$ = 55.5
Weight of water=1000g ( $∵$ 1000 mL=1000 g)
Mole fraction = 10^-5 = $\frac{x}{55.5 + x}$

As 55.5 >>> x, thus neglecting x from denominator
$10^{- 5} = \frac{x}{55 .5} \Rightarrow x = 55 .5 \times 10^{- 5} moles$
or 0.555 millimoles.

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