Question

Henry’s law constant for $C{O}_2$ in water is $1.67\times {10}^{8}Pa$at 298K the quantity of $C{O}_2$ in 500 ml of soda water when packed under 2.5atm pressure is:
$$\begin{gathered} \text{A}\text{. 0}\text{.082 mole} \\ \text{B}\text{. 0}\text{.82 mole} \\ \text{C}\text{. 0}\text{.41 mole} \\ \text{D}\text{. 0}\text{.042 mole} \end{gathered}$$

Answer 1

Hint: In this question first we find the mole fraction of $C{O}_2$ with the help of henry’s law. Then using the formula of mole fraction and applying the approximation on it to get the number of moles of $C{O}_2$.

Formula used:
Henry’s law: $\text{p = }{\text{K}}_H\times x$; $$x_{solvent}={\displaystyle \frac{n_{solvent}}{n_{solvent}+n_{solute}}}$$

Complete answer:
We know, according to Henry's law, the mass of a dissolved gas in a given volume of solvent at equilibrium is proportional to the partial pressure of the gas.
$$\Rightarrow \text{p = }{\text{K}}_H\times x$$ (Equation 1)
Where,
p= partial pressure of gas
$${\text{K}}_H$$= Henry’s constant$(1.67\times {10}^{8}Pa)$
x= mole fraction of dissolved gas
Given: pressure of $C{O}_2$=2.5atm
We know 1 atm=$1.01325\times {10}^{5}Pa$
Then,
Pressure of $C{O}_2$=$2.5\times 1.01325\times {10}^{5}Pa$
$=2.533125\times {10}^{5}Pa$
Now, on applying the Henry’s law on $C{O}_2$ we get
$$\begin{gathered} \Rightarrow \text{p = }{\text{K}}_H\times x \\ \Rightarrow x={\displaystyle \frac{\text{p}}{{\text{K}}_H}} \end{gathered}$$
On putting the values in above equation, we get
$$\begin{gathered} \Rightarrow x={\displaystyle \frac{2.533125\times {10}^{5}Pa}{1.67\times {10}^{8}Pa}} \\ \Rightarrow x=1.52\times {10}^3 \end{gathered}$$
Therefore the mole fraction of $C{O}_2$ is $1.52\times {10}^3$.
But we have 500ml of soda water so that
Volume of water=500ml
Density of water=1g/ml
We know, mass=volume$\times$density
And 500 ml of water=500g of water
Molar fraction of water=18g/mol
Now, the number of moles of water=${\displaystyle \frac{\text{Mass of water}}{\text{Molar mass of water}}}$
Then,
The number of moles of water=${\displaystyle \frac{500}{18}}$
$=27.78mol$
Now, apply the formula of mole fraction in $C{O}_2$, we get
$$\Rightarrow x={\displaystyle \frac{n_{c{o}_2}}{n_{c{o}_2}+n_{H_{2}O}}}$$
Since,
So,
$$\Rightarrow x={\displaystyle \frac{n_{c{o}_2}}{n_{H_{2}O}}}$$
On putting the values which we have in above equation, we get
$$\begin{gathered} \Rightarrow n_{c{o}_2}=27.78\times 1.52\times {10}^{-3} \\ \Rightarrow n_{c{o}_2}=0.042mole \end{gathered}$$
Therefore, the moles of $C{O}_2$ is 0.042mole.

Hence, option D is correct.

Note: Whenever you get this type of question the key concept to solve this is to learn the concept of Henry’s law , its formula $$\text{p = }{\text{K}}_H\times x$$ and mole fraction and its formula $$x_{solvent}={\displaystyle \frac{n_{solvent}}{n_{solvent}+n_{solute}}}$$. And one more thing to be noted is that density of water is 1g/ml so mass of water is equal to volume of water.