Question

a gas has a henrys law constant of 7.80×10⁻⁴ m/atm at 18.3 °c. what volume of solution is needed to completely dissolve 1.97 l of the gas at 1235 torr and 18.3 °c?

Explanation:

Step1: Convert pressure from torr to atm

1 atm = 760 torr. So, $P=\frac{1235}{760}\ atm\approx1.625\ atm$.

Step2: Use the ideal - gas law ($PV = nRT$) to find the number of moles of the gas

At constant temperature ($T = 18.3^{\circ}C=(18.3 + 273.15)K=291.45K$), $n=\frac{PV}{RT}$. Assuming $R = 0.0821\ L\cdot atm/(mol\cdot K)$, $n=\frac{1.625\ atm\times1.97\ L}{0.0821\ L\cdot atm/(mol\cdot K)\times291.45K}$.
$n=\frac{3.20125}{23.927045}mol\approx0.134\ mol$.

Step3: Use Henry's law ($C = kP$) to find the concentration of the gas in the solution

Given $k = 7.80\times10^{-4}\ M/atm$ and $P = 1.625\ atm$, $C=kP=(7.80\times10^{-4}\ M/atm)\times1.625\ atm = 1.2675\times10^{-3}\ M$.

Step4: Calculate the volume of the solution

Since $C=\frac{n}{V}$, then $V=\frac{n}{C}$. Substituting $n = 0.134\ mol$ and $C = 1.2675\times10^{-3}\ M$, $V=\frac{0.134\ mol}{1.2675\times10^{-3}\ mol/L}\approx106\ L$.

Answer:

106 L