Question

what mass of a non volatile solute urea needs to be dissolved in hundred gram of water in order to decrease the vapour pressure of water by 30% what will be the molality of solution​

Answer 1

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Answer 2

Explanation:

By Raoult's law,

$\sf \dfrac{P^{\circ} - P_S}{P_S}=\dfrac{W_2Mw_1}{W_1Mw_2}$

If,

• $\sf P^{\circ} = 100 \: nm$
• $\sf P_S = 70 \: nm$

$\sf \dfrac{100-70}{70}=\dfrac{W_2×18}{100×60}$

$\sf W_2=143.33$

$\sf Molality = \dfrac{W_2×1000}{Mw_2×W_1}$

$\sf = \dfrac{143.33×1000}{60×100}$

$\sf = 23.88$