The molarity of 1 L orthophosphoric acid having 70% purity by weight (specific gravity 1.54 g ) is ______ M

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Solution

Q.
The molarity of 1 L orthophosphoric acid ( $H_{3} {PO}_{4}$ ) having 70% purity by weight (specific gravity 1.54 g ${cm}^{- 3}$ ) is ______ M.
(Molar mass of $H_{3} {PO}_{4}$ = 98 g ${mol}^{- 1}$ ) [2024]

Ans.
(11)

   $70% w/w H_{3} {PO}_{4} solution means:$

   $70 g H_{3} {PO}_{4} in 100 g solution$

$\Rightarrow \frac{70}{98} mol H_{3} {PO}_{4} in \frac{100}{1.54} mL solution$

$M = \frac{n_{solute}}{V_{solution} (in mL)} \times 1000$

   $M = \frac{70 / 98}{100 / 1.54} \times 1000 M = \frac{70 \times 1.54 \times 1000}{98 \times 100} M = 11 M$

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