Question

The vapor pressure of pure liquid $C{S}_2$is$0.3914atm$at$20°c$. Whenof rhombic sulfur is dissolved in$1.00kg$of$C{S}_2$, the vapor pressure of$C{S}_2$decreases to$0.3914atm$. Determine the molecular formula for the sulfur molecules dissolved in$C{S}_2$.

Short answer

The molecular formula of rhombic sulfur is${\mathrm{S}}_8$ .

Step-by-step solution

Given Information

The vapor pressure of pure liquid ${\mathrm{CS}}_2$is $0.3914a\mathrm{tm}|$. But after the addition of rhombic sulfur, the vapor pressure decreases to role="math" localid="1663402342740" $0.3868a\mathrm{tm}$.

Concept of Raoult’s law

The vapor pressure of a solvent above a solution equals the vapor pressure of a pure solvent at the same temperature scaled by the mole fraction of the solvent present $\left({\mathrm{X}}_{\mathbf{solvent}}\right)$.

According to Raoult's law:role="math" localid="1663402438789" ${\mathbf{P}}_{\mathbf{solution}}\mathbf{}\mathbf{=}{\mathrm{X}}_{\mathbf{solvent}}{{\mathbf{P}}^{\mathbf{0}}}_{\mathbf{solvent}}$

Calculation of the molar fractions

The dissolved rhombic sulphur in carbon disulphide constitute a molar fraction, which can be calculated as follows:

$$\begin{gathered} X_1=\frac{\frac{1.00\times {10}^3\text{g}}{76.1407\text{g}/\text{mol}}}{\frac{1.00\times {10}^3\text{g}}{76.1407\text{g}/\text{mol}}+n_{R.S}} \\ {X}_1=\frac{13.1}{13.1+n_{RS}} \\ 0.3868\text{atm}=\frac{131}{131+n_{ns}}0.3914\text{atm} \\ \frac{13.1}{13.1+n_{R.S}}=\frac{0.3868\text{atm}}{0.3914\text{atm}}=0.9882 \end{gathered}$$

Calculate the mole of sulphur

Substitute the value from above step and calculate further:

$$\begin{gathered} 13.1=0.9882\left(13.1+n_{RS}\right) \\ =12.9+0.9882n_{RS} \\ 0.9882n_{Rs}=13.1-12.945 \\ =0.2945 \end{gathered}$$

Further steps in continuation

$$\begin{gathered} n_{RS}=\frac{0.155}{0.9882} \\ =0.157 \end{gathered}$$

Calculation of molar mass

Substitute the value in formula:

$$\begin{gathered} \text{Molar mass}=\frac{\text{Mass}}{\text{Number of moles}} \\ =\frac{40.0\text{g}}{0.157\text{mol}} \\ =254.8\text{g}/\text{mol} \\ \approx 255\text{g}/\text{mol} \end{gathered}$$

Calculate the molecular formula

Thus, the ratio of molar mass to atomic mass gives the mole of sulphur.

$$\begin{gathered} \frac{255\text{g}/\text{mol}}{32.065\text{g}/\text{mol}}=7.95 \\ \approx 8 \end{gathered}$$