Answer to Question #165646 in Chemistry for Agacer

Question

 Calculate the weight  of water in a 1.5 liters 95% nitric acid (HNO3) that has a specific gravity  of 1.82.  Express the concentration of solution in terms of Molarity, Normality and molality.        

Answer 1

1) density = m/V

Mass of HNO₃ = density x V = 1.82 x 1.5 = 2.73 kg

% = (100 x m_(solvent))/ (m_(solvent) + m_(solute))

m(HNO₃) = 95/100 x 2.73 = 2.59 kg

m(H₂O) = 2.73 - 2.59 = 0.14 kg

2) C_M = n/V

M (HNO₃) = 63.01 g/mol

n (HNO₃) = m/M = 2590/63.01 = 41.10 mol

С_M = 41.10/1.5 = 27.4 mol/L

3) C_n = N/V

N = n/equivalence factor

n (HNO₃) = 41.10 mol

Equivalence factor (HNO₃) = 1

N (HNO₃) = 41.10/1 = 41.10

C_n (HNO₃) = 41.10/1.5 = 27.4 eq/L

4) C_m = n_(solute)/m_(solvent)

C_m (HNO₃) = 41.10/0.14 = 292.93 mol/kg