Question

The molarity of HNO₃ in a sample which has density 1.4 g/mL and mass percentage of 63% is _____. (Molecular Weight of HNO₃ = 63)

Answer 1

100 gm soln → 63 gm HNO₃

100/1.4mL → 1 mole HNO₃

Molarity = 1/(100/1.4 x 1/1000) = 14M

Answer 2

M = (63 x 1.4 x 10)/63 = 14