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 $ $mm$ $Hg$
(b) $0.031 $ $mm$ $Hg$
(c) $0.017 $ $mm$ $Hg$
(d) $0.028 $ $mm$ $Hg$
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Answer:
Correct Answer: 28. $(180.40$ $ mL)$
Solution:
Let us consider $1.0 L$ of solution.
$\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 \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=5.412 $
$\Rightarrow 2.706 \times 100 =270.6 V $
$ V =180.40$ $ mL$
BETA
