Question

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

• A. $0.555$
• B. $55.55\times {10}^{-5}$
• C. $55.55\times {10}^{-3}$
• D. $5.55\times {10}^{-5}$

Answer 1

Answer: (B)

$55.55\times {10}^{-5}$

Given:-

Partial pressure of gas, p = 1bar, $K_H$ = 100kbar = 100×1000 bar.

p = $K_H$ × mole fraction.

So, mole fraction = ${10}^{-}5$

Weight of water $=1L=1000ml=1000g,$

molecular weight of $H_{2}O$$=18\frac{g}{mol}$

Moles of water = $\frac{wt.}{mol.wt.}$ = $\frac{1000}{18}=55.5$

Let $x=$ mole of gas dissolved.

Mole fraction, ${10}^{-5}$ $=\frac{x}{55.5+x}$

Since x is very small as compared to 55.5, it is neglected from the denominator.

Now,

${10}^{-5}$ = $\frac{x}{55.5}$

$x=55.5\times {10}^{-}5$.