The molarity of 1L 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 = 98g mol ^ - 1 )

Correct answer: 0 Specific gravity (density) = 1 .54 g / cc . Volume = 1L = 1000ml Mass of solution = 1 .54 × 1000 = 1540g % purity of H _ 3 PO _ 4 is 70 % So weight of H _ 3 PO _ 4 = 0 .7 × 1540 = 1078g Mole of H _ 3 PO _ 4 = (mass)/(molarmass) = (1078)/(98) = 11 Molarity= (No . ofmolesofsolute)/(VolumeofsolutioninL) Molarity = (11)/(1L) = 11M

JEE Main 202431 JanuaryEvening Shift

Exam: JEE Main
Year: 2024
Date: 31 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Main 2024 · 31 January · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

The molarity of 1L orthophosphoric acid ( H _ 3 PO _ 4 ) having 70...

The molarity of $1L$ orthophosphoric acid $\left({H}_{3}{PO}_{4}\right)$ having $70 %$ purity by weight (specific gravity $1 .54 g{cm}^{- 3}$ ) is ______ $M$ . (Molar mass of ${H}_{3}{PO}_{4} = 98g{mol}^{- 1}$ )

Correct answer

Step-by-step solution

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

Volume $= 1L = 1000ml$

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

$= 1540g$

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

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

Mole of ${H}_{3}{PO}_{4} = \dfrac{mass}{molarmass} = \dfrac{1078}{98} = 11$

Molarity= $\dfrac{No . ofmolesofsolute}{VolumeofsolutioninL}$

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

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