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:
$
  {\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}} \\
$

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: ${\text{p = }}{{\text{K}}_H} \times x$; \[{x_{solvent}} = \dfrac{{{n_{solvent}}}}{{{n_{solvent}} + {n_{solute}}}}\]

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
$
   \Rightarrow {\text{p = }}{{\text{K}}_H} \times x \\
   \Rightarrow x = \dfrac{{\text{p}}}{{{{\text{K}}_H}}} \\
 $
On putting the values in above equation, we get
$
   \Rightarrow x = \dfrac{{2.533125 \times {{10}^5}Pa}}{{1.67 \times {{10}^8}Pa}} \\
   \Rightarrow x = 1.52 \times {10^3} \\
 $
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 $density
And 500 ml of water=500g of water
Molar fraction of water=18g/mol
Now, the number of moles of water=$\dfrac{{{\text{Mass of water}}}}{{{\text{Molar mass of water}}}}$
Then,
The number of moles of water=$\dfrac{{500}}{{18}}$
$=27.78mol$
Now, apply the formula of mole fraction in $C{O_2}$, we get
\[ \Rightarrow x = \dfrac{{{n_{c{o_2}}}}}{{{n_{c{o_2}}} + {n_{{H_2}O}}}}\]
Since,
So,
\[ \Rightarrow x = \dfrac{{{n_{c{o_2}}}}}{{{n_{{H_2}O}}}}\]
On putting the values which we have in above equation, we get
$
   \Rightarrow {n_{c{o_2}}} = 27.78 \times 1.52 \times {10^{ - 3}} \\
   \Rightarrow {n_{c{o_2}}} = 0.042mole \\
$
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 \[{x_{solvent}} = \dfrac{{{n_{solvent}}}}{{{n_{solvent}} + {n_{solute}}}}\]. 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.