4 g of non-volatile solute dissolved in 36 g of water produces the relative lowering of vapour pressure of 0.1 The molecular mass of the solute is

Question

4 g of non-volatile solute dissolved in 36 g of water produces the relative lowering of vapour pressure of 0.1 The molecular mass of the solute is (A) 40 (B) 20 (C) 80 (D) 100

Tardigrade Admin · Accepted Answer

(p° -ps/p°) =(w2 M1/w1M2) 0.1= (4 ×18/36× M2) M2 = (4× 18/36× 0.1) =20

Q. 4 g of non-volatile solute dissolved in 36 g of water produces the relative lowering of vapour pressure of 0.1 The molecular mass of the solute is

40

20

80

100

$\frac{p ^{\circ} - p _{s}}{p ^{\circ}} = \frac{w _{2} M _{1}}{w _{1} M _{2}}$
$0.1 = \frac{4 \times 18}{36 \times M _{2}}$
$M_{2} = \frac{4 \times 18}{36 \times 0.1} = 20$

Q. 4 g of non-volatile solute dissolved in 36 g of water produces the relative lowering of vapour pressure of 0.1 The molecular mass of the solute is

40

20

80

100

p∘p∘−ps​​=w1​M2​w2​M1​​ 0.1=36×M2​4×18​ M2​=36×0.14×18​=20

$\frac{p ^{\circ} - p _{s}}{p ^{\circ}} = \frac{w _{2} M _{1}}{w _{1} M _{2}}$
$0.1 = \frac{4 \times 18}{36 \times M _{2}}$
$M_{2} = \frac{4 \times 18}{36 \times 0.1} = 20$