Single Choice

Stock phosphoric acid solution is $$85$$% $$H_3PO_4$$ and has a specific gravity of $$1.70$$ . Hence, molarity of $$H_3PO_4$$ solution is:

A$$8.51$$ M
B$$1.70$$ M
C$$14.74$$ M
Correct Answer
D$$7.37$$ M

Solution

$$100$$ g of phosphoric acid solution contains $$85$$ g of $$H_3PO_4$$.
The specific gravity is $$1.7$$ g/mL.
Hence, $$\displaystyle \dfrac {100}{1.7} mL\space H_3PO_4$$ contains $$85$$ g $$H_3PO_4$$. It is equal to $$\displaystyle \dfrac {85}{98}$$ moles of $$H_3PO_4$$.

$$Molarity = \displaystyle \dfrac {85/98 mol}{ \bigg ( \dfrac {100\ L}{1.7 \times 1000} \bigg )}=14.74 M$$.

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