Molarity of a 50 mL `H_2SO_4` solution is 10.0 M. If the density of the solution is 1.4 g/cc, calculate its molality

Options

A. 7.14
B. 8
C. 10
D. 0.5

Explanation

Given , molarity of `H_2SO_4` solution = 10.0 M
Volume (V) = 50 mL = .05 L
Density of solution (d) = 1.4 g//cc
Molarity 10m means 1 mole of `H_2SO_4` present in 1L (1000 mL) of solution
also , molar mass of `H_2SO_4 (M_n) = 98`
`because` 1000 mL has `H_2SO_4 = 980 g (98 xx 10)`
`therefore` 50 mL has `H_2SO_4 = (980 xx 50)/(1000) = 49 g`
thus, mass of solute `(W_B) = 49 g`
Calculation for mass of solution and solvent:
` because` d = mass of solution / volume of solution
`therefore ` mass of solution ` = d xx V = 1.4 xx 50 = 70 g`
Thus mass of solvent `(W_A)` = mass of solution - mass of solute . = 70 - 49
` W_A = 21 g`
Calculation for molarity (m):
`m = (W_B)/(M_B) xx (1000)/(W_(AKg)) = 49/98 xx 1000/21 = 23.80 m`
molarity (m) = 23.8 m