Question

A sample of dry nitrogen gas weighing 100.0g is bubbled through liquid water at 25.0⁰C. The gaseous mixture of nitrogen and water vapor escapes at a total pressure of 700.0 torr. What mass of water has vaporized? (The vapor pressure of water at 25.0⁰C is 23.8 torr.)

Short answer

The mass of water that has vaporized is equal to 2.26 g.

Step-by-step solution

Determine the moles of dry nitrogen.

The number ofmoles present in

$$\begin{gathered} n_{N_2}=\frac{\text{mass}}{\text{molecular mass}} \\ {n}_{N_2}=\frac{100.0g}{2\times 14.01g/mol} \\ {n}_{N_2}=3.5689mol \end{gathered}$$

Determine the moles of water.

For ideal gas: $pV=nRT$

The number of moles present in water:

$$\begin{gathered} \frac{p_{H_{2}O}}{p_T}=\frac{n_{H_{2}O}RT/V}{n_{T}RT/V} \\ \frac{p_{H_{2}O}}{p_T}=\frac{n_{H_{2}O}}{n_{H_{2}O}+n_{N_2}} \\ {n}_{H_{2}O}=\frac{p_{H_{2}O}\times n_{N_2}}{p_T-p_{H_{2}O}} \\ {n}_{H_{2}O}=\frac{23.8torr\times 3.57mol}{700torr-23.8torr} \end{gathered}$$

By solving, we get$n_{H_{2}O}=0.126mol$

Determine the mass of water that has vaporized.

$$\begin{gathered} The mass of water=n_{H_{2}O}\times \text{Molecular mass} \\ =0.1256mol\times 18.02g/mol \\ =2.26g \end{gathered}$$

Hence the mass of water is 2.26 g.