At room temperature, a dilute solution of urea is prepared by...

Relative lowering in vapour Pressure
$⇒$ Mole of urea \left({n}_{B}\right) = \dfrac{0.6}{60} = {10}^{- 2}mole$;</p><p>Mole of water \left({n}_{A}\right) = \dfrac{360}{18} = 20$

$\dfrac{{P}^{o} - {P}_{s}}{{P}^{o}} = {x}_{B} = \dfrac{{P}^{o} - {P}_{s}}{{P}_{o}} = \dfrac{{n}_{B}}{{n}_{n} + {n}_{B}}$

Here $= {P}^{o} = V . P .$ of pure solvent

${P}_{s} = V . P$ of solution.

${n}_{A} + {n}_{B} ≃ {n}_{A}$

$\dfrac{{P}^{o} - {P}_{s}}{{P}^{o}} = \dfrac{{n}_{B}}{{n}_{A}}$
$loweringofvapourpressure ( {P}^{o} - {P}_{s} ) = \dfrac{{n}_{B}}{{n}_{A}} \times {P}^{o}$

${P}^{o} - {P}_{s} = \dfrac{{10}^{- 2}}{20} \times 35$

${P}^{o} - {P}_{s} = 0.0175mm of Hg$