At 25$^0$C, the saturated solution of a salt in water was found to contain 0.24g of the salt in 100cm$^3$. What is the solubility of the salt in gdm$^{-1}$?

Analyze the question: The solubility is given as 0.24 g of salt in 100 cm³ of solution. You are asked for it in g/dm³ (the "gdm⁻¹" in the question is a typo for g dm⁻³).

Concept: Solubility in g/dm³ is the mass of solute (in grams) dissolved in 1 dm³ of solution.

1 dm³ = 1000 cm³

Solution:

• 100 cm³ contains 0.24 g
• 1000 cm³ will contain (0.24 × 1000) / 100 g

= 240 / 100

= 2.40 g/dm³

Check: 1000 cm³ is 10 times 100 cm³, so the mass must be 10 times larger: 0.24 × 10 = 2.40 g. ✔

Answer: Option C, 2.40 g/dm³.

Tip: To convert from g/100 cm³ to g/dm³, just multiply by 10 (move the decimal point one place to the right). Common mistakes are multiplying by 100 (which gives 24.0, option D) or dividing by 10 (which gives 0.024, option A).