Question

When gas is bubbled through water$\text{298K}$, 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, the number of moles of gas dissolved in 1 litre of water is :
A) $0.555$
B) $55.55\times {{10}^{\text{ -}5}}$
C) $55.55\times {{10}^{\text{-}3}}$
D) $5.55\times {{10}^{-5}}$

Answer 1

The number of moles of gas dissolved in the given volume of solution can be calculated by Henry’s law. According to which the partial pressure p of the gas dissolved in the solution is directly related to the mole fraction x of the gas and has a proportionality constant ${{\text{K}}_{\text{H}}}$. The mole fraction for a mixture of components is given as, number of moles of the gas per total number of moles present in the system.

Complete Step by step solution:
The pressure on gas dissolved per unit volume of the solution is proportional to the mole fraction of the dissolved substance. This is called Henry’s law. Thus maybe also expressed as
$\text{p=}{{\text{K}}_{\text{H}}}x$
Where x is the mole fraction of gas dissolved in solution, ${{\text{K}}_{\text{H}}}$is Henry's constant, and p is the partial pressure of the gas.
We are given with the following data,
${{\text{K}}_{\text{H}}}\text{=100Kbar=100 }\\!\\!\times\\!\\!\text{ 1000bar=1}{{\text{0}}^{\text{5}}}\text{bar}$
Partial pressure,$\text{p=1 bar}$
Let us rearrange the in Henry’s law concerning mole fraction
$\text{x=}\dfrac{\text{Partial pressure on gas}}{{{\text{K}}_{\text{H}}}}$
$\text{x=}\dfrac{\text{1 bar}}{{{10}^{5}}\text{bar}}$
Therefore, mole fraction is $\text{x=1}{{\text{0}}^{\text{-5}}}$
We have to find the number of moles that dissolve in the water. Let’s first find the moles of water. To do so let us find out the weight of water.
We are provided with the 1liter of the solution.
Thus the weight of water is equal to the $\text{1L=1000mL=1000g}$
The molecular weight of water ${{\text{M}}_{{{\text{H}}_{\text{2}}}\text{O}}}=18\text{ g/mol}$
Thus the number of moles of water calculated as:
$\text{No}\text{.of moles = }\dfrac{\text{mass of }{{\text{H}}_{\text{2}}}\text{O}}{\text{mol}\text{. wt}\text{.of }{{\text{H}}_{\text{2}}}\text{O}}$
Substitute the values.
$\text{No}\text{.of moles = }\dfrac{\text{1000 g}}{\text{18 g/mol}}=55.5\text{ mol}$
Thus the number of moles of water for one-litre solutions is$55.5\text{ mol}$.
Now let x be the number of moles of gas dissolved in the one-litre solution.
But we know that the mole fraction for a mixture is always equal to one.
Therefore,
${{\text{x}}_{\text{gas}}}\text{=}\dfrac{{{\text{n}}_{\text{gas}}}}{{{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{+}{{\text{n}}_{\text{gas}}}}$
Substitute the values. We get,
$\text{1}{{\text{0}}^{\text{-5}}}\text{=}\dfrac{\text{x}}{\text{55}\text{.5+x}}$
Since x is very small than the 55.5 $\text{(x}\langle \langle \langle \text{55}\text{.5)}$ we can neglect it from the denominator.
$\text{1}{{\text{0}}^{\text{-5}}}\text{=}\dfrac{\text{x}}{\text{55}\text{.5}}$
Or $\text{x=55}\text{.5}\times \text{1}{{\text{0}}^{\text{-5}}}$
The mole fraction for gas is $\text{55}\text{.5}\times \text{1}{{\text{0}}^{\text{-5}}}$ which can be also written as the $0.555\text{ millimoles}$

Hence, (B) is the correct option.

Note: Henry’s law is applicable only for the gas which is not largely soluble even though the solution is saturated or maybe dilute.
The density of water is$1\text{ Kg/}{{\text{m}}^{\text{3}}}$. Therefore the volume of water is directly considered as the mass of water. Therefore here,$\text{1000mL=1000g}$.