Chemistry › Solutions › Henry's law
At 293 K , methane gas was passed into 1 L of water. The partial pressure of methane is 1 bar . The number…
At 293 K , methane gas was passed into 1 L of water. The partial pressure of methane is 1 bar . The number of moles of methane dissolved in 1 L water is $\left(\mathrm{K}_{\mathrm{H}}\right.$ of methane $=0.4 \mathrm{kbar}$ )
$1.38$ $1.38 \times 10^{-2}$ $1.38 \times 10^{-3}$ $1.38 \times 10^{-1}$ Solution
According to Henry's Law,
$\begin{aligned} & \mathrm{p}=\mathrm{K}_{\mathrm{H}} \times \mathrm{x} \\ & \mathrm{p}=\text { partial pressure } \\ & \mathrm{K}_{\mathrm{H}}=\text { Henery constant }\end{aligned}$
$\mathrm{x}=$ mole fraction
given
$\mathrm{p}=1 \text { bar }$
$\begin{aligned} & \mathrm{K}_{\mathrm{H}}=0.4 \times 10^3 \text { bar }\left[1 \mathrm{~K} \text { bar }=10^3 \text { bar }\right] \\ & \therefore \mathrm{p}=\mathrm{K}_{\mathrm{H}} \mathrm{x}\end{aligned}$
$\begin{aligned} & 1 \text { bar }=0.4 \times 10^3 \text { bar } \times x \\ & x=\frac{1}{0.4 \times 10^3}\end{aligned}$
$\begin{aligned} & \frac{\mathrm{n}_{\text {methane }}}{\mathrm{n}_{\text {methane }}+\mathrm{nH}_2 \mathrm{O}}=\frac{1}{0.4 \times 10^3} \\ & \mathrm{n}_{\text {methane }}\lt \lt \mathrm{nH}_2 \mathrm{O}\end{aligned}$
$\begin{aligned} & \therefore \frac{\mathrm{n}_{\text {methane }}}{\mathrm{nH}_2 \mathrm{O}}=\frac{1}{0.4 \times 10^3} \\ & \frac{\mathrm{n}_{\text {methane }}}{\mathrm{nH}_2 \mathrm{O}}=\frac{1}{0.4 \times 10^3} \quad\left[\mathrm{nH}_2 \mathrm{O}=\frac{1000}{18}=55.55 \mathrm{~mol}\right]\end{aligned}$
$\begin{aligned} & =\frac{55.55}{0.4 \times 10^3} \\ & =138 \times 10^{-1}\end{aligned}$
$\mathrm{n}_{\text {methane }}=1.38 \times 0 \times 10^{-1}$
Asked in: AP EAMCET 2024 (18 May Shift 1)
Practice more Solutions questions on Aicharya