Question

If the solubility of carbon monoxide in water at $0^{o}C$ and 1 atm pressure as 0.0354 mL CO per mL of solution. What should be the pressure of $CO\left(g\right)$ above the solution to obtain $0.010MCO$ solution?

• A. 0.633 atm
• B. 6.33 atm
• C. 1 atm
• D. 0.01 atm

Answer 1

The correct option is B 6.33 atm

$\text{Volume of}CO=0.0354mL/mL\text{solution}=0.0354\times {10}^{-3}L/mL\text{solution}$
P= 1 atm, T = 273 K
$\therefore n_{\left(CO\right)}=\frac{pV}{RT}$
$=\frac{1atm\times 0.0354\times {10}^{-3}L/mL\text{solution}}{0.0821Latmmo{l}^{-1}{K}^{-1}\times 273K}$

$=1.5794\times {10}^{-6}mol/mL\text{solution}$

$\text{In}1000mLor1L\text{solution,number of moles of CO are:}$

$\Rightarrow \frac{1.5794\times {10}^{-6}mol}{0.001L}$

$=1.5794\times {10}^{-3}M$

By Henry's law, solubility of a gas inside the solution is directly proportional to its pressure above the solution at constant temperature.
$p_1=K_H\times C_1{p}_2=K_H\times C_2$
$\therefore \frac{C_2}{C_1}=\frac{p_2}{p_1}$
$p_2=\frac{C_2}{C_1}\times p_1=\frac{0.01M\times 1.0atm}{1.5794\times {10}^{-3}M}$
$p_2=6.33atm$