Vapour pressure of a solution of 5g of non- electrolyte in 100g of water at a particular temperature is 2985N/m2 . The vapour pressure of pure water is 3000N/m2 . The molecular weight of the solute is

Question

Vapour pressure of a solution of 5g of non- electrolyte in 100g of water at a particular temperature is 2985N/m2 . The vapour pressure of pure water is 3000N/m2 . The molecular weight of the solute is (A) 60 (B) 120 (C) 180 (D) 380

Tardigrade Admin · Accepted Answer

(Po - Ps/Po)=((W2/M2)/(W1)M1) =( text3000 - text2985/ text3000)=((5/M2)/( text100) text18) text or M2= text180

Q. Vapour pressure of a solution of $5 g$ of non- electrolyte in $100 g$ of water at a particular temperature is $2985 N / m^{2}$ . The vapour pressure of pure water is $3000 N / m^{2}$ . The molecular weight of the solute is

60

120

180

380

$\frac{P ^{o} - P _{s}}{P ^{o}} = \frac{\frac{W _{2}}{M _{2}}}{\frac{W _{1}}{M _{1}}}$
$= \frac{3000 - 2985}{3000} = \frac{\frac{5}{M _{2}}}{\frac{100}{18}} or M_{2} = 180$

Q. Vapour pressure of a solution of 5g of non- electrolyte in 100g of water at a particular temperature is 2985N/m2 . The vapour pressure of pure water is 3000N/m2 . The molecular weight of the solute is

60

120

180

380

PoPo−Ps​​=M1​W1​​M2​W2​​​ =30003000−2985​=18100​M2​5​​ or M2​=180

Q. Vapour pressure of a solution of $5 g$ of non- electrolyte in $100 g$ of water at a particular temperature is $2985 N / m^{2}$ . The vapour pressure of pure water is $3000 N / m^{2}$ . The molecular weight of the solute is

$\frac{P ^{o} - P _{s}}{P ^{o}} = \frac{\frac{W _{2}}{M _{2}}}{\frac{W _{1}}{M _{1}}}$
$= \frac{3000 - 2985}{3000} = \frac{\frac{5}{M _{2}}}{\frac{100}{18}} or M_{2} = 180$