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Solutions — Chemistry STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceChemistrySolutions5 Marks
Question
  1. State Raoult’s law for a solution containing volatile components.
 How does Raoult’s law become a special case of Henry’s law?
  1. $1.00\ g$ of a non-electrolyte solute dissolved in $50\ g$ of benzene lowered the freezing point of benzene by $0.40 K.$ Find the molar mass of the solute. $(K_f$ for benzene $= 5.12 \ kg\ mol^{-1}).$
✓

Answer

  1. Partial vapour pressure of a liquid component is directly proportional to its mole fraction in its solution.
The partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. Only the proportionality constant $K_H$ differs from $P^o_A.$ Thus, Raoult’s law becomes a special case of Henry’s law in which $K_H$ becomes equal to $P^o_A.$
  1. Given $W_B = 1∙00g;$
 $W_A = 50g;$ $K_f = 5∙12K \ kg/\ mol- 1;$   $\triangle T_f = 0∙40K$
 $\Delta\text{T}_{f}=\text{K}_{f}\frac{\text{W}_{B}\times{1000}}{\text{M}_{B}\times\text{W}_{A}\text{(in grams)}}$
$\text{M}_{B}=\text{K}_{f}\frac{\text{W}_{B}\times{1000}}{\Delta\text{T}_{f}\times\text{W}_{A}}$
$\text{M}_{B}=\frac{\text{5.12}\times{1}\times{1000}}{\text{0.40}\times\text{50}}$
$= 256\ g\ mol^{-1}$​​​​​​​

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