Question

20% (w/v) aqueous solution of H2SO4 has density 1.2 gm/mL, then molality (m)
of solution will be
1.02 m
2.04 m
(1)
(2)
(3)
(4)
3.5 m
4.20 m​

Answer 1

Given : Molarity of H
2
​
SO
4
​
= 18M

As, Molarity =
volume of solution
no. of moles of solute
​

i.e., 18moles of sulphuric acid is present per litre of solution.

Density=1.8g/mL {given}
Thus , mass of one litre of this solution ,
Density =
volume
mass
​

mass = density×volume
mass = 1.8×1000
mass = 1800g
Thus , mass of 1litre 18M H
2
​
SO
4
​
solution is 1800g or 1.8kg.
Now, mass of solution = mass of solute + mass of solvent
mass of solution = no. of moles of solute X molar mass of solute + mass of solvent
1800 = 18×98 + mass of solvent
mass of solvent = (1800−1764)g
mass of solvent = 36g
Now , molality of H
2
​
SO
4
​
solution is given by,

molality =
massofsolvent(inkg)
numberofmolesofsolute
​


molality =
36
18×100
​

molality = 500m

Hence, the molality of H
2
​
SO
4
​
solution is 500m.