Question

ON DISSOLVING 3.24g OF SULPHUR IN 40g OF BENZENE, BOILING POINTOF THE SOLUTION WAS HIGHER THAN THAT OF BENZENE BY 0.81K, Kb VALUE FOR BENZENE IS 2.53 K Kg mol​-1 . What IS THE MOLECULAR FORMULA OF SULPHUR?

Answer 1

Answer: 253 g/mol

Explanation: Formula used for Elevation in boiling point :

$\Delta T_b=k_b\times m$

or,

$T_{solution}-T_{solvent}=k_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

where,

$T_b$ = change in boiling point  = 0.81 K

$k_b$ = boiling point constant  = 2.53 K/kgmol

m = molality

weight of solvent (benzene) = 40 g= 0.04 kg (1kg= 1000g)

mass of solute (sulphur)= 3.24 g

Molar mass of solute (sulphur)= M g/mol

$0.81=2.53\times \frac{3.24g}{Mg/mol\times 0.04kg}$

$M=253g/mol$

Answer 2

Answer:

$S_8$ is the molecular formula of the sulfur dissolved.

Explanation:

Elevation in boiling point is determined by formula :

Let the molecular formula of sulfur be $S_x$ amd molar mass of the same be M.

$\Delta T_b=k_b\times m$

$\Delta T_b=T_b-T$

where,

$\Delta T_b$= change in boiling point  = 0.81 K

$T_b$=Boiling point of the solution

T = boiling point of the pure solvent

$k_b$ = boiling point constant  = 2.53 K/kgmol

m = molality

Weight of solvent (benzene) = 40 g = 0.04 kg (1kg= 1000g)

mass of solute (sulphur)= 3.24 g

Molar mass of solute (sulphur)= M g/mol

$0.81=2.53\times \frac{3.24g}{Mg/mol\times 0.04kg}$

M=253g/mol

Atomic mass of single sulfur atom is 32 g/mol

Molar mass of $S_x$ is 253 g/mol

$32 g/mol\times x=253 g/mol$

$x=\frac{253 g/mol}{32 g/mol}=7.9 \approx 8$

$S_8$ is the molecular formula of the sulfur dissolved.