Mass of Urea ( NH _ 2 CONH _ 2 ) required to be dissolved in 1000g...

Relative lowering of vapour pressure=Mole fraction of solute

$\dfrac{Loweringofvapourpressure}{Vapourepressureofsolution} = \dfrac{molesofsolute}{molesofsolvent}$

$\dfrac{25}{75} = \dfrac{\text{moles of urea}}{\left(\dfrac{1000}{18}\right)}$

$\dfrac{1}{3} \times \dfrac{1000}{18} =$moles of urea

Mass of urea$= \dfrac{1}{3} \times \dfrac{1000}{18} \times 60$

$= 1111 . 11gm$