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Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2024 — 31 January, Shift 2 — Question 77

The molarity of 1 L1\ \mathrm{L} orthophosphoric acid (H3PO4)\mathrm{(H_3PO_4)} having 70%70\% purity by weight (specific gravity 1.54 g cm−31.54\ \mathrm{g\ cm^{-3}}) is _____\_\_\_\_\_ MM.

[Given: molar mass of H3PO4=98 g mol−1\mathrm{H_3PO_4} = 98\ \mathrm{g\ mol^{-1}}]

Answer: 11

Numerical answer — enter this value.

Step-by-step solution

Specific gravity (density) =1.54 g/cc=1.54 \mathrm{~g} / \mathrm{cc}.

Volume =1 L=1000ml=1 \mathrm{~L}=1000 \mathrm{ml}

Mass of solution =1.54×1000=1.54 \times 1000 =1540  g =1540\; \mathrm{g}

% purity of H2SO4\mathrm{H}_{2} \mathrm{SO}_{4} is 70%70 \%

So, mass of pure H3PO4=0.7×1540=1078 g\mathrm{H}_{3} \mathrm{PO}_{4}=0.7 \times 1540=1078 \mathrm{~g}

Moles of H3PO4=107898=11\mathrm{H}_{3} \mathrm{PO}_{4}=\frac{1078}{98}=11

Molarity =111 L=11=\frac{11}{1 \mathrm{~L}}=11

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic
Concentration Terms and Their Interconversion

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