Question

3.765g of sulphur is dissolved in 100g of $C{{S}_{2}}$. This solution boils at 319.81K. What is the molecular formula of sulphur in solution? The boiling point of $C{{S}_{2}}$ is 319.45K. (Given that ${{K}_{b}}$ for $C{{S}_{2}}$ = 2.42m k Kg $mo{{l}^{-1}}$ and atomic mass S = 32?

Answer 1

Hint: Firstly writing the given part may help us to avoid much confusion, and gives us a better picture of the question. To answer this question we should know the formula to calculate molar mass of the given solute.

Complete step by step solution:
According to this question, sulphur is the solute and $C{{S}_{2}}$ is the solvent.
Given,
The atomic mass of sulphur = 32
The mass of the sulphur = 3.765g
= 3.76 $\times {{10}^{-3}}$Kg
The mass of the $C{{S}_{2}}$ = 100g
= 100 $\times {{10}^{-3}}$ Kg
The boiling point of the solution (${{T}_{b}}$ ) = 319.81 K
The boiling point of solvent (${{T}_{b}}^{\circ }$) = 319.45K
Molar elevation constant (${{K}_{b}}$) = 2.42m k Kg $mo{{l}^{-1}}$
Firstly, let's $\Delta {{T}_{b}}$,
\[\Delta {{T}_{b}}={{T}_{b}}-{{T}_{b}}^{\circ }\]
$\implies 319.81 - 319.45$
$\therefore 0.36$
Let's calculate the molar mass of sulphur,
The formula to calculate molar mass is
\[Molar\text{ }mass\text{ }of\text{ }solute=\dfrac{{{K}_{b}}\times Weight\text{ }of\text{ }solute}{\Delta {{T}_{b}}\times Weight\text{ }of\text{ }solvent}\]
The molar mass of sulphur :
\[Molar\text{ }mass\text{ }of\text{ }sulphur=\dfrac{{{K}_{b}}\times Weight\text{ }of\text{ sulphur}}{\Delta {{T}_{b}}\times Weight\text{ }of\text{ C}{{\text{S}}_{2}}}\]
$\implies$ \[\dfrac{2.42\times 3.795\times {{10}^{-3}}}{0.36\times 100\times {{10}^{-3}}}\]
$\implies$ 0.2551 Kg $mo{{l}^{-1}}$
$\therefore$ 255.1g $mo{{l}^{-1}}$
Therefore, molar mass of sulphur is 255.1g $mo{{l}^{-1}}$
Now, let's calculate the number of atom of sulphur in $C{{S}_{2}}$:
\[Number\text{ }of\text{ }moles\text{ }of\text{ }sulphur=\dfrac{Molar\text{ }mass\text{ }of\text{ }sulphur\times X}{Atomic\text{ }mass\text{ }of\text{ }sulphur~~~~~~~~~~}\]
 The number of moles is 1 ,
Let X be number of sulphur atom
\[1=\dfrac{255.1\times X}{32}\]
$X = 7.97$
$X = 8$ (approximately)

Thus, the molecular formula of sulphur in $C{{S}_{2}}$ is ${{S}_{8}}$.

Note: Firstly, we should know to figure out the solute and solvent in the given question. Secondly, writing may give us a better picture to solve this question. It is better to write the unit and in the above solution the mass of sulphur and $C{{S}_{2}}$ is changed from gram to kilogram in order to make calculation easy.