Commercially available concentrated hydrochloric acid contains $38$ $H C l$ by mass.
$(i)$ What is the molarity of the solution (density of solution = $1.19 g m L^{- 1}$ )?
$(i i)$ What volume of concentrated $H C l$ is required to make $1.0 L$ of an $0.10 M H C l ?$

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Solution

$38$ $g m$ $H C l$ is present in $100$ $g m$ solution.
Volume of solution $= \frac{100}{1.19} = 84$ $m L$
Moles of $H C l = \frac{38}{36.5} = 1.04$ $m o l e s$
$(i)$ $M o l a r i t y = \frac{m o l e s}{V o l u m e o f s o l u t i o n (l i t r e s)} = \frac{}{1.04 \times 1000} 84 = 12.38$ $M$
$(i i)$ No. of moles of $H C l$ require $= 0.1$ $m o l e s$
Mass of $H C l = 0.1 \times 36.5 = 3.65$ $g m$
Mass of solution having $3.65$ $g m$ $H C l = y$
$\frac{38}{100} = \frac{3.65}{y}$
$⟹ y = 9.6$ $g m$
Volume of concentration $H C l$ required $= \frac{9.6}{1.19} = 8.067$ $m L$

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