Question

What is the correct formula for solubility of a solute in a solvent at any given temperature t⁰C?

• A. $\frac{\text{Weight of solute(Kg)}\ast 100}{\text{Weight of solution}\left(Kg\right)}$
• B. $\frac{\text{Weight of solvent}\left(g\right)\ast 100}{\text{Weight of solute(g)}}$
• C. $\frac{\text{Weight of solute(g)}}{\text{Weight of solvent(g)}\ast 100}$
• D. $\frac{\text{Weight of solute(g)}\ast 100}{\text{Weight of solvent(g)}}$

Answer 1

The correct option is D $\frac{\text{Weight of solute(g)}\ast 100}{\text{Weight of solvent(g)}}$

Let at t⁰C, 'a' grams of solvent dissolves maximum 'b' grams of solute to produce a saturated solution, then the solubility of that solute at t⁰C is

=$\frac{\text{Weight of solute(g)}}{\text{Weight of solvent(g)}}\times 100$ = $\frac{b}{a}\times 100$