The molarity (M) of an aqueous solution containing 5.85\ g of NaCl in 500\ mL water is: (Given : Molar mass of Na = 23 gmol -1 and Cl = 35.5 gmol -1 )

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

JEE Main 2024 — 4 April, Shift 1 — Question 71

The molarity $(M)$ of an aqueous solution containing $5.85\ \mathrm{g}$ of $\mathrm{NaCl}$ in $500\ \mathrm{mL}$ water is:

(Given : Molar mass of $\mathrm{Na} = 23 \mathrm{gmol}^{-1}$ and $\mathrm{Cl} = 35.5 \mathrm{gmol}^{-1}$ )

Molarity formula, $\mathrm{M = \frac{n}{V}}$

Molar mass of $\mathrm{NaCl}$ $= \mathrm{23 + 35.5 = 58.5\ g\ mol^{-1}}$

Moles $= \mathrm{\frac{5.85}{58.5} = 0.1\ mol}$

Volume $= \mathrm{500\ mL = 0.5\ L}$

Molarity $= \mathrm{M = \frac{0.1}{0.5} = 0.2}$

Answer key and solution verified before publishing.

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