Solutions: Real-World Scenarios (4)

easy 2 min read

Question

A doctor recommends a saline drip with 0.9%0.9\% (w/v) NaCl. Calculate the molarity of this solution. (NEET 2022 pattern)

Solution — Step by Step

0.9%0.9\% (w/v) means 0.9 g0.9\ \text{g} of NaCl per 100 mL100\ \text{mL} of solution. Equivalently, 9 g9\ \text{g} per litre.

Molar mass of NaCl =23+35.5=58.5 g/mol= 23 + 35.5 = 58.5\ \text{g/mol}.

moles=958.50.154 mol per litre\text{moles} = \frac{9}{58.5} \approx 0.154\ \text{mol per litre}
M=nV=0.154 mol1 L=0.154 M0.15 MM = \frac{n}{V} = \frac{0.154\ \text{mol}}{1\ \text{L}} = 0.154\ \text{M} \approx 0.15\ \text{M}

Final answer: Molarity 0.15 M\approx 0.15\ \text{M} (also called isotonic saline, matching plasma osmolarity).

Why This Works

Concentration units are just bookkeeping: w/v percent says “grams per 100 mL”, molarity says “moles per litre”. Convert the units consistently, and the answer falls out.

The reason saline drips use this exact concentration is that 0.15 M0.15\ \text{M} NaCl is approximately isotonic with blood plasma, preventing red blood cells from shrivelling or bursting.

Alternative Method

Use molality if density is given. For dilute aqueous solutions, molality \approx molarity. For concentrated solutions, the difference matters and we use m=nsolutekg solventm = \tfrac{n_{\text{solute}}}{\text{kg solvent}}.

For “ppm” (parts per million), 1 ppm = 1 mg per litre of dilute aqueous solution. JEE Main 2023 had a fluoride-in-water PYQ using this.

Common Mistake

Confusing w/v and w/w. 0.9%0.9\% (w/v) is grams per 100 mL of solution. 0.9%0.9\% (w/w) is grams per 100 g of solution — slightly different because solutions are denser than water. Read the question carefully.

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