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
