Question

A handbook lists the solubility of carbon monoxide in water at $0^{\circ }C$ and $1atm$ pressure as $0.0410mLCO$ per mL of $H_{2}O$. What should be the pressure of $CO\left(g\right)$ above the solution to obtain $0.030MCO$ solution?

• A. $0.633atm$
• B. $6.33atm$
• C. $4.93mm$
• D. $16.41atm$

Answer 1

The correct option is D $16.41atm$

$\text{Volume of}CO=0.0410mL=0.0410\times {10}^{-3}L$

$\text{p = 1 atm, T = 273 K}$

$\therefore n_{CO}=\frac{pV}{RT}$

$=\frac{1\text{atm}\times 0.0410\times {10}^{-3}L}{0.0821{\text{L atm mol}}^{-1}{\text{K}}^{-1}\times 273K}$


$=1.8276\times {10}^{-6}\text{mol}$
This amount of $CO$ is dissolved in $1ml$ of water. So, molarity$\left(C\right)=$ number of mol in $1000ml=1.8276\times {10}^{-6}\times 1000=1.8276\times {10}^{-3}M$
Using Henry's law, $p=k_{H}x,k_H=\frac{1atm}{1.8276\times {10}^{-3}M}$
Now, solubility needed=$0.03M$, $k_H$ remains constant at a particular temperature.
From Henry's law, $p=k_{H}x=\frac{1atm}{1.8276\times {10}^{-3}M}\times 0.03M=16.41atm$
Hence, (d) is correct.