Chemistry - Concentration Terms Question with Solution | TestHub

ChemistryConcentration TermsConcentration termMedium2 minPYQ_2024

ChemistryMediumassertion reason

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 $H_{3} {PO}_{4} = 98 g {mol}^{- 1}$ )

Answer:

Solution:

Specific gravity (density) $= 1 .54 g / cc$ .

Volume $= 1 L = 1000 ml$

Mass of solution $= 1 .54 \times 1000$

$= 1540 g$

$%$ purity of $H_{3} {PO}_{4}$ is $70 %$

So weight of $H_{3} {PO}_{4} = 0 .7 \times 1540 = 1078 g$

Mole of $H_{3} {PO}_{4} = \frac{mass}{molar mass} = \frac{1078}{98} = 11$

Molarity= $\frac{No . of moles of solute}{Volume of solution in L}$

Molarity $= \frac{11}{1 L} = 11 M$

Stream:JEESubject:ChemistryTopic:Concentration TermsSubtopic:Concentration term

⏱ 2mℹ️ Source: PYQ_2024

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