SolutionHard
Question
A quantity of 6.4 g sulphur dissolved in 200 g of CS2 raises the boiling point of the solvent by 0.32°C. Boiling point of CS2 = 47°C and latent heat of vaporization of CS2 = 80.0 cal/g. The molecular formula of sulphur in this liquid is
Options
A.S8
B.S4
C.S2
D.S
Solution
$K_{b} = \frac{0.002T^{2}}{\Delta H_{\text{vap}}\left( \text{Cal/gm} \right)} = \frac{0.002 \times (320)^{2}}{80} = 2.56\text{ K/m}$
Now, $\Delta T_{b} = K_{b}.m \Rightarrow 0.32 = 2.56 \times \frac{6.4/32x}{200/1000} \Rightarrow x = 8$
∴ Molecular formula of sulphur = Sx = S8
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