Question

The boiling point of a solution of $5g$ of sulphur in $100g$ of carbon disulphide is ${0.476}^{\circ }C$ above that of pure solvent. Determine the molecular formula of sulphur in this solvent. The boiling point of pure carbon disulphide is ${46.30}^{\circ }C$ and its heat of vapourisation is $84.1cal/g$

• A. $S_4$
• B. $S_{12}$
• C. $S_8$
• D. $S_6$

Answer 1

The correct option is C $S_8$

We know,
Boiling point elevation constant,
$K_b=\frac{R{T}_b^2}{1000L_v}$
where,
$T_b$ is boiling point of pure solvent in $K$
$R\text{in}cal/mol/K$
$L_v$ is latent heat of vaourisation of solvent in $cal/g$
Thus,
$K_b=\frac{2\times (46.30+273{)}^2}{1000\times 84.1}=2.425$
Boiling point elevation of a solution is given by,
$△T_b=K_b\times m$
$\therefore m=\frac{△T_b}{K_b}=\frac{0.476}{2.425}$
And also, $m=\frac{\text{moles of solute}}{\text{wt. of solvent in grams}}\times 1000$
$=\frac{5}{M}\times \frac{1000}{100}=\frac{50}{M}$
where, M is the molar weight of sulphur.
Thus, $\frac{50}{M}=\frac{0.476}{2.425};M=255$
Since, the atomic weight of sulphur is 32 and its molar weight is 255.
Therefore, number of sulphur atoms associated to form a single molecule is $\frac{255}{32}=7.96$ i.e., 8.
Hence the formula is $S_8.$