JEE Main 2025 Jan 28 Shift 1, Chemistry Q23: Types Concentration And Solubility of Solutions

JEE Main2025Jan 28, Shift 1Chemistry

The molarity of a 70% (mass/mass) aqueous solution of a monobasic acid (X) is _______ M (Nearest integer)

[Given : Density of aqueous solution of (X) is 1.25 g mL $^{- 1}$
Molar mass of the acid is 70 g mol $^{- 1}$ ]

(Express your answer as the integer $\times 1 0^{- 1}$ as printed in the original key.)

Your answer

Solution

Take 100 g of solution as a basis.

Mass of solute (acid X) = 70 g, so moles of X = $70/70 = 1$ mol.

Volume of 100 g of solution $= \frac{100}{1.25} = 80$ mL = 0.080 L.

$M = \frac{mol solute}{L solution} = \frac{1}{0.080} = 12.5$ M.

Reported as $\times 1 0^{- 1}$ : $12.5 = 125 \times 1 0^{- 1}$ .

$∴$ Answer = $125$ .

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