12 g of a nonvolatile solute dissolved in 108 g of water produces the relative lowering of vapour pressure of 0.1. The molecular mass of the solute is :

Question

12 g of a nonvolatile solute dissolved in 108 g of water produces the relative lowering of vapour pressure of 0.1. The molecular mass of the solute is : (A) 80 (B) 60 (C) 20 (D) 40

Tardigrade Admin · Accepted Answer

(Po-Ps/Po) = (n/N) = (w/m)× (M/W) 0.1= (12/m) × (18/108) m = (12×18/0.1×108) = 20

Q. $12 g$ of a nonvolatile solute dissolved in $108 g$ of water produces the relative lowering of vapour pressure of $0.1.$ The molecular mass of the solute is :

80

60

20

40

$\frac{P ^{o} - P _{s}}{P ^{o}} = \frac{n}{N} = \frac{w}{m} \times \frac{M}{W}$
$0.1 = \frac{12}{m} \times \frac{18}{108}$
$m = \frac{12 \times 18}{0.1 \times 108} = 20$

Q. 12g of a nonvolatile solute dissolved in 108g of water produces the relative lowering of vapour pressure of 0.1. The molecular mass of the solute is :

80

60

20

40

PoPo−Ps​​=Nn​=mw​×WM​ 0.1=m12​×10818​ m=0.1×10812×18​=20

Q. $12 g$ of a nonvolatile solute dissolved in $108 g$ of water produces the relative lowering of vapour pressure of $0.1.$ The molecular mass of the solute is :

$\frac{P ^{o} - P _{s}}{P ^{o}} = \frac{n}{N} = \frac{w}{m} \times \frac{M}{W}$
$0.1 = \frac{12}{m} \times \frac{18}{108}$
$m = \frac{12 \times 18}{0.1 \times 108} = 20$