Question

A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm^–3. The molarity of the solution is ...... mol dm^–3. (Round off to the Nearest Integer). [Atomic masses: K :39.0 u; O :16.0 u; H :1.0 u]

Answer 1

By using the relation,

\(m = \frac{1000 \times M}{(1000 \times d) - M \times M_{solute }}\)

Where, m = molality of the solution = 6.50 m

M = molarity of the solution

d = density = 1.89 g cm⁻³

M_solute = Molar mass of solute = 39 + 16 + 1 = 56 u

Substituting the values, we get

\(6.5 = \frac{1000 \times M}{1890 - M \times 56}\)

⇒ 12285 - 364 M = 1000 M

⇒ 1364 M = 12285

⇒ M = 9 mol dm^-3

Answer 2

6.5 molal KOH = 1000gm solvent has 6.5 moles KOH

so wt of solute = 6.5 × 56

= 364 gm

wt of solution = 1000 + 364 = 1364