Search any question & find its solution
Question:
Answered & Verified by Expert
Calculate the quantity of $\mathrm{CO}_2$ required to prepare $1 \mathrm{~L}$ of soda water when the soda water was packed under $2 \mathrm{~atm}$ of $\mathrm{CO}_2$. [Henry's law constant for $\mathrm{CO}_2$ is $1.67 \times 10^8 \mathrm{~Pa}$ ]
Options:
Solution:
2719 Upvotes
Verified Answer
The correct answer is:
2.9 g
According to Henry's law;
$$
\begin{aligned}
& \mathrm{x}=\frac{\mathrm{P}}{\mathrm{K}_{\mathrm{H}}}=\frac{2 \times 1.013 \times 10^5}{1.67 \times 10^8} \\
& \frac{\mathrm{n}_{\mathrm{co}_2}}{1000 / 18}=\frac{2 \times 1.013}{1.67 \times 1000} \\
& \mathrm{n}_{\mathrm{co}_2}=\frac{2 \times 1.013}{1.67 \times 18} \Rightarrow \text { moss of } \mathrm{CO}_2 \text { dissolved } \\
& =\frac{2 \times 1.013 \times 44}{1.67 \times 18}=2.9 \mathrm{~g}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{x}=\frac{\mathrm{P}}{\mathrm{K}_{\mathrm{H}}}=\frac{2 \times 1.013 \times 10^5}{1.67 \times 10^8} \\
& \frac{\mathrm{n}_{\mathrm{co}_2}}{1000 / 18}=\frac{2 \times 1.013}{1.67 \times 1000} \\
& \mathrm{n}_{\mathrm{co}_2}=\frac{2 \times 1.013}{1.67 \times 18} \Rightarrow \text { moss of } \mathrm{CO}_2 \text { dissolved } \\
& =\frac{2 \times 1.013 \times 44}{1.67 \times 18}=2.9 \mathrm{~g}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.