Concentrations of solutions
5 marks · HardA student is given 3.90 g of ammonium chloride (NH₄Cl, Ar: N = 14, H = 1, Cl = 35.5) and asked to make 250 cm³ of a solution.
Calculate the amount, in mol, of NH₄Cl in 3.90 g. (Give your answer to 2 decimal places.)
Calculate the concentration, in mol/dm³, of the NH₄Cl solution prepared.
State the concentration in g/dm³ for this solution.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) 0.07 mol
(b) 0.28 mol/dm³
(c) 15.6 g/dm³
Common mistakes: Omitting to convert cm³ to dm³; Incorrect calculation of molar mass
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correctly calculates molar mass of NH₄Cl (14 + (1×4) + 35.5 = 53.5 g/mol) or states it.
A1 (a) · 1 mark · Correctly calculates amount of NH₄Cl: 3.90 g ÷ 53.5 g/mol = 0.0729 mol (accept 0.07 mol to 2 decimal places).
M2 (b) · 1 mark · Converts 250 cm³ to 0.250 dm³ and applies concentration = amount / volume.
A2 (b) · 1 mark · Calculates concentration: 0.0729 mol / 0.250 dm³ = 0.292 mol/dm³ (accept 0.29 or 0.28 mol/dm³ rounded). (follow-through)
B1 (c) · 1 mark · States concentration as 15.6 g/dm³ (3.90 g in 0.25 dm³ × 4 = 15.6 g/dm³). (follow-through)
Worked solution
a) Molar mass of NH₄Cl = 14 + (1×4) + 35.5 = 53.5 g/mol
b) Volume in dm³: 250 cm³ = 0.250 dm³
c) In 0.250 dm³, there are 3.90 g; in 1 dm³, there will be 3.90 × 4 = 15.6 g/dm³.