The density of 1.40 molal solution of acetic acid `(CH_3COOH)` is 1.084 g Ml`""^(-1).` The molarity of solution will be a)1.30M b)1.40M c)1.50M d)1.60M

Options

A. 1.30M
B. 1.40M
C. 1.50M
D. 1.60M

Answer

B

Explanation

Let the mass of solvent be 1kg. We will have `m_(1)= 1000g and n_(2)=1.40mol`
Mass of acetic acid, `m_(2) = n_(2) M_(2)= (1.40 mol) (60g mol^(-1))===84g`
Mass of solution, `m= m_(1) m_(2)= 1000g + 84g = 1084g`
Volume of solution, `V= (m)/(rho)= (1084g)/(1.084g mL^(-1))= 1000 mL = 1L`
Molarity of solution `M = (n_(2))/(V)= (1.40 mol)/(1L)= 1.40M`