Solutions and Colligative Properties - Result Question 29
####28. What is the molarity and molality of a $13 %$ solution (by weight) of sulphuric acid with a density of $1.02 g / mL$ ? To what volume should $100 mL$ of this solution be diluted in order to prepare a $1.5 N$ solution?
$(1978,2 M)$
(a) $0.027 mmHg$
(b) $0.031 mmHg$
(c) $0.017 mmHg$
(d) $0.028 mmHg$
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Solution:
- Let us consider $1.0 L$ of solution.
$$ \begin{array}{rlrl} \text { Weight of solution } & =1000 \times 1.02=1020 g \ \text { Weight of } H _2 SO _4 & =1020 \times \frac{13}{100}=132.60 g \ \text { Weight of } H _2 O & =1020-132.60=887.40 g \ \Rightarrow \quad & \text { Molarity } & =\frac{132.60}{98}=1.353 M \ & \text { Molality } & =\frac{132.60}{98} \times \frac{1000}{887.40}=1.525 m \ \Rightarrow & & \text { Normality } & =2 \times M=2.706 \ \Rightarrow & & 2.706 \times 100 & =1.5 V \ & & V & =180.40 mL \end{array} $$
