Question

The solubility of $N_2$ in water is $2.2\times {10}^{-4}g$ in $100g$ of $H_{2}O$ at ${20}^{\circ }C$ when the pressure of $N_2$ over the solution is $1.2atm$.

Calculate the solubility at ${20}^{\circ }C$ when the pressure of $N_2$ over the solution is $10atm$.

• A. $1.25\times {10}^{-3}g/100g{H}_{2}O$
• B. $1.49\times {10}^{-3}g/100g{H}_{2}O$
• C. $2.56\times {10}^{-3}g/100g{H}_{2}O$
• D. $1.83\times {10}^{-3}g/100g{H}_{2}O$

Answer 1

The correct option is D $1.83\times {10}^{-3}g/100g{H}_{2}O$

According to Henry's law, $S=KP$
S is the solubility in moles per litre
K is the henry's law constant
P is the pressure in atm.
For 1.2 atm pressure, substitute values in the above expression.
$2.2\times {10}^{-4}g/100g{H}_{2}O=K\times 1.2atm$
$K=1.833\times {10}^{-4}g/100g{H}_{2}Oat{m}^{-1}$
For 10 atm pressure, substitute values in the above expression.
$S=1.833\times {10}^{-4}g/100g{H}_{2}Oat{m}^{-1}\times 10atm=1.83\times {10}^{-3}g/100g{H}_{2}O$.