Concentrated sulfuric acid contains very little water, only 5.0% by mass. it has a density of 1.84g/ml. what is the molarity of this acid

When m for water is only 5 % so, m for the acid is (m%) = 95%

and D = 1.84 g/ml, assume 1 L of solution therefore,

1 - D = m / v

m = D x v = 1.84 g/ml x 1 L x 10^3 ml / L

= 1840 g solution

2 - m% = m acid / m sol x 100 %

m acid = 1840 g H2So4 x 0.95

= 1750 g H2So4

3 - n = 1750 g H2So4 x 1 mol H2So4 x 98.09 H2So4

= 17.8 mol H2So4

4 - So the molarity of { H2So4} = n / v = 17.8 mol H2So4 / 1 L

= 17.8 M