JEE MAIN - Chemistry (2025 - 28th January Morning Shift - No. 23)
The molarity of a $70 \%$ (mass/mass) aqueous solution of a monobasic acid (X) is _________ $\times 10^{-1}$ M (Nearest integer)
[Given: Density of aqueous solution of (X) is $1.25 \mathrm{~g} \mathrm{~mL}^{-1}$
Molar mass of the acid is $70 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
Answer
125
Explanation
Assuming 100 gm solution contain 70 gm solute.
Volume of 100 gm solution will be $\frac{100}{1.25} \mathrm{ml}$.
Molarity $=\frac{70 / 70}{100 / 1.25} \times 1000=12.5$ or $125 \times 10^{-1}$
Comments (0)
