At 27°C, 58.5g of sodium chloride is present in 250cm ^3 of a solution. The solubility of sodium chloride at…

Analyze the question: We need the solubility in mol/dm³. This means the number of moles of NaCl dissolved per dm³ of solution.

Concept: Solubility (mol/dm³) = (mass ÷ molar mass) ÷ volume in dm³

Solution using the molar mass given in the question (111.0 g/mol):

1. Moles = 58.5 ÷ 111.0 = 0.527 mol
2. Volume = 250 cm³ = 0.25 dm³
3. Solubility = 0.527 ÷ 0.25 ≈ 2.1 mol/dm³

This is closest to A (2.0 mol/dm³).

Note on the data: The real molar mass of NaCl is 58.5 g/mol (Na = 23, Cl = 35.5). Using that, 58.5 g is exactly 1 mole, so the solubility would be 1 ÷ 0.25 = 4.0 mol/dm³. This is not among the options. The value 111.0 g/mol in the question is actually the molar mass of CaCl₂, so the question has an error. Using the values as printed in the question, the answer that fits is about 2.1, so option A (2.0 mol/dm³) is the intended one.

Answer: A. 2.0 mol/dm³

Exam tip: Always convert cm³ to dm³ by dividing by 1000 before calculating. In an exam, use the molar mass the question gives you, even if it looks unusual. Pick the option closest to your result.