Question

The quantity of $CO _{2}$ in $500\, mL$ of soda water when packed under $3.34$ bar $CO _{2}$ pressure at $298\, K$ in $g$ is

• A. 2.442
• B. 1.221
• C. 4.884
• D. 3.663

Answer 1

Answer: (A)

Given, Quantity of $CO _{2}=500\, mL$

Pressure of $CO _{2}= 3.34$ bar

Temperature $=298\, K$ in Henry's law

$p=K_{ H } X \times(x)$

$X=\frac{p}{K_{ H }}=3.34\, bar =334000\, Pa$

$=\frac{3.34 \times 10^{1}}{1.67 \times 10^{8}}=2.442\, Pa$

But, we have $500\, mL$ of soda water so that

Volume of water $=500\, mL$

Density of water $= 1\, g / mL$

Formula mass $=$ volume $\times$ density we get

$500\, mL$ of water $=500 \,g$ of water

Molar mass of water $=18\, g / mol ^{-1}$

$\frac{500}{18}=27.78$ mole $\mu$ of water

$x=\frac{n_{ CO _{2}}}{n_{ CO _{2}}+n_{ H _{2} O }}$

Value of moles fraction is very small so it is negligible as compared to 1 we get,

$x=\frac{n_{ CO _{2}}}{n_{ H _{2} O }}=2 \times 10^{-7}$

Molar mass of $CO _{2}$ is after calculation $=2.442\, g$