The quantity of $\mathrm{CO}_2$ in $500 \mathrm{~mL}$ of soda water when packed under 3.34 bar…
The quantity of $\mathrm{CO}_2$ in $500 \mathrm{~mL}$ of soda water when packed under 3.34 bar $\mathrm{CO}_2$ pressure at $298 \mathrm{~K}$ in $\mathrm{g}$ is
2.442
1.221
4.884
3.663
Solution
Given,
Quantity of $\mathrm{CO}_2=500 \mathrm{~mL}$
Pressure of $\mathrm{CO}_2=3.34 \mathrm{bar}$
Temperature $=298 \mathrm{~K}$ in
Henry's law
$
\begin{gathered}
p=K_{\mathrm{H}} X \times(x) \\
X=\frac{p}{K_{\mathrm{H}}}=3.34 \text { bar }=334000 \mathrm{~Pa}
\end{gathered}
$
$
=\frac{3.34 \times 10^1}{1.67 \times 10^8}=2.442 \mathrm{~Pa}
$
But, we have $500 \mathrm{~mL}$ of soda water so that Volume of water $=500 \mathrm{~mL}$
Density of water $=1 \mathrm{~g} / \mathrm{mL}$
Formula mass $=$ volume $\times$ density we get $500 \mathrm{~mL}$ of water $=500 \mathrm{~g}$ of water
Molar mass of water $=18 \mathrm{~g} / \mathrm{mol}^{-1}$
$
\begin{aligned}
\frac{500}{18} & =27.78 \text { mole } \mu \text { of water } \\
x & =\frac{n_{\mathrm{CO}_2}}{n_{\mathrm{CO}_2}+n_{\mathrm{H}_2 \mathrm{O}}}
\end{aligned}
$
Value of moles fraction is very small so it is negligible as compared to 1 we get,
$
x=\frac{n_{\mathrm{CO}_2}}{n_{\mathrm{H}_2 \mathrm{O}}}=2 \times 10^{-7}
$
Molar mass of $\mathrm{CO}_2$ is after calculation $=2.442 \mathrm{~g}$