Question

the density of a solution containing 13% sulphuric acid in1.09g/ml . then the molarity of solution is

Answer 1

13% by mass means 13 g of H2SO4 in 100 g of solution 

molecular mass of H2SO4 = 98 g/mole 

so no.of moles of H2SO4 = 13/98 = 0.133 

and volume of solution = mass of solution/density = 100/1.09 = 91.743 ml = 0.092 L 

molarity = 0.133/0.092 = 1.446 moles/L 

normality = molarity X n-factor = molarity X 2 = 1.446 X 2 = 2.892 N 

n-factor for an acid = no.of ionizable H+ ....in case of H2SO4 = 2 

Answer 2

Answer: 1.44 M

Explanation: Density= 1.09

1.09= mass of solution/volume of solution

When density is 1.09, there is 109 g of solution in 100 ml of solution or 1090 g in 1000 ml of solution.

Mass of solution= 1090 g

Mass of H2SO4= 13% of 1090 g

= (13/100)*1090

= 1417/10

= 141.7 g

Molar mass of H2SO4= 98 g

Molarity= (m/M)*(1/V)*1000

= (141.1/98)*(1/1000)*1000

= 1.44 M