Given that 10 g of a dibasic acid (mol. mass=100) are present in 500 mL of the solution. The density of the solution is 1.02 g ml. Match the entries of column I with appropriate entries of column II and choose the correct option.

Options

A. A-p, B-2, C-r, D-s
B. A-s, B-r, C-9, D-p
C. A-s, B-r, C-p, D-2
D. A-s, B-r, C-p, D-2

Answer

B

Explanation

Normality ` = (10)/(50 ) xx (1)/( 500 ) xx 1000 = 0.4`
Mass of solution `= 500 xx 1.02 = 510 g`
Mass of solvent `= 510 - 10 g = 500 g`
` therefore ` Molality`= (10)/(100 ) xx (1)/(500 ) xx 1000 = 0.2`
` 500 g H_2 O = 500 // 18 mol = 27.78 mol`
10 g acid `=(10)/(100) ` mol = 0.1 mol
Mole fraction `: X_(H_2 O) = ( 27.78 )/( 27.78 +0.1 )= 0.996 `
Mass fraction `= 500 // 510 = 0.98`