Single Choice

Molarity of $$H_2 SO_4$$ (density 1.8 g / mL) is 18 M. The molality of this $$H_2 SO_4$$ solution is:

A$$36$$
B$$200$$
C$$500$$
Correct Answer
D$$18$$

Solution

$$Given$$ : Molarity of $$H_2SO_4$$ = $$18 M$$

As, Molarity = $$\dfrac{no. \ of \ moles \ of \ solute}{volume \ of \ solution}$$
i.e., $$18 moles$$ of sulphuric acid is present per litre of solution.

$$Density = 1.8 g/mL $$ {given}
Thus , mass of one litre of this solution ,
Density = $$\dfrac{mass}{volume} $$
mass = $${density \times volume }$$
mass = $${1.8 \times 1000} $$
mass = $$1800 g $$
Thus , mass of $$1 litre$$ $$18 M$$ $$H_2SO_4$$ solution is $$1800 g$$ or $$1.8 kg$$.
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
$$1800 $$ = $${18 \times 98}$$ + mass of solvent
mass of solvent = $$(1800 - 1764)g $$
mass of solvent = $$36 g $$
Now , molality of $$H_2SO_4$$ solution is given by,

molality = $$\dfrac{number of moles of solute}{mass of solvent(in kg)}$$

molality = $$\dfrac{18 \times 100 }{36}$$
molality = $$500 m $$

Hence, the molality of $$H_2SO_4$$ solution is $$500 m $$.

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