Lowering of vapour pressure of $1.00 \mathrm{~m}$ solution of a non-volatile solute in a hypothetical…

Lowering of vapour pressure of $1.00 \mathrm{~m}$ solution of a non-volatile solute in a hypothetical solvent of molar mass $40 \mathrm{~g}$ at its normal boiling point, is :
  1. $29.23$ torr
  2. $30.4$ torr
  3. $35.00$ torr
  4. $40.00$ torr

Solution

$\mathrm{P}_{\text {solvent }}^{0}=760$ torr $($ at the b. pt.) $\mathrm{n}_{\text {solute }}=1$
$\mathrm{n}_{\text {solvent }}=\frac{1000}{40}=25$
$\Delta \mathrm{P}=760 \times \mathrm{X}_{\text {solute }}=\frac{760}{26}$ torr $=29.23$ torr .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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