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If $\mathrm{CO}_2$ gas having a partial pressure of 1.67 bar is bubbled through l L water at $298 \mathrm{~K}$, the amount of $\mathrm{CO}_2$ dissolved in water in $\mathrm{g} \mathrm{L}^{-1}$ is approximately. (Henry's law constant of $\mathrm{CO}_2$ is $1.67 \mathrm{k}$ bar at $298 \mathrm{~K}$ )
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Verified Answer
The correct answer is:
2.44
Let $p_A$ be the partial pressure of $\mathrm{CO}_2$ and $\chi_A$ be its mole fraction
$$
p_A=K_{\mathrm{H}} \chi_A(\text { Henry's law) }
$$
$1.67 \mathrm{bar}=1.67 \mathrm{k}$ bar $\times \chi_A$
(where, $\quad x_A=0.001$ )
or
$$
\frac{n_A}{n_B}=0.001
$$
$[\because$ Number of moles of water $=55$ ]
$$
\begin{aligned}
n_A & =0.001 \times 55.55 \\
n_A & =0.056
\end{aligned}
$$
$$
\begin{aligned}
& \quad \quad n=\frac{\text { Given mass }(m)}{\text { Molecular mass }(M)}=0.056 \\
& \therefore \quad m=0.056 \times 44=2.44 \mathrm{~g} \\
& \text { Thus, } 2.44 \mathrm{~g} \text { of } \mathrm{CO}_2 \text { is dissolved in } 1 \mathrm{~L} \text { water at } \\
& 298 \mathrm{~K} \text {. }
\end{aligned}
$$
$298 \mathrm{~K}$.
$$
p_A=K_{\mathrm{H}} \chi_A(\text { Henry's law) }
$$
$1.67 \mathrm{bar}=1.67 \mathrm{k}$ bar $\times \chi_A$
(where, $\quad x_A=0.001$ )
or
$$
\frac{n_A}{n_B}=0.001
$$
$[\because$ Number of moles of water $=55$ ]
$$
\begin{aligned}
n_A & =0.001 \times 55.55 \\
n_A & =0.056
\end{aligned}
$$
$$
\begin{aligned}
& \quad \quad n=\frac{\text { Given mass }(m)}{\text { Molecular mass }(M)}=0.056 \\
& \therefore \quad m=0.056 \times 44=2.44 \mathrm{~g} \\
& \text { Thus, } 2.44 \mathrm{~g} \text { of } \mathrm{CO}_2 \text { is dissolved in } 1 \mathrm{~L} \text { water at } \\
& 298 \mathrm{~K} \text {. }
\end{aligned}
$$
$298 \mathrm{~K}$.
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