The molarity of a 70% (mass /mass) aqueous solution of a monobasic acid (X) is __________ $$\times10^{-1}$$ M(Nearest integer) [Given: Density of aqueous solution of (X) is 1.25 g $$mL^{-1}$$ Molar mass of the acid is $$70 g mol^{-1}$$]

Find the molarity of a 70% (mass/mass) aqueous solution of a monobasic acid with density 1.25 g/mL and molar mass 70 g/mol.

Assuming 100 g of solution, the mass of acid is 70% of 100 g = 70 g, and the mass of water is 100 − 70 = 30 g.

Using the density, the volume of the solution is given by:

$$ V = \frac{\text{mass}}{\text{density}} = \frac{100 \text{ g}}{1.25 \text{ g/mL}} = 80 \text{ mL} = 0.080 \text{ L} $$

The number of moles of acid is

$$ n = \frac{70}{70} = 1 \text{ mol} $$

Thus the molarity is

$$ M = \frac{n}{V} = \frac{1}{0.080} = 12.5 \text{ M} $$

Expressing this in the required form gives $$12.5 = 125 \times 10^{-1}$$ M.

The answer is 125.