A 75 gm 4% w/w aqueous NaOH solution is distilled for some time. The water collected has a mass of 25 gms. To this distilled solution X gms of KOH ( s ) is added, so that molality of OH − ion in the r

Question

A 75 gm 4% w/w aqueous NaOH solution is distilled for some time. The water collected has a mass of 25 gms. To this distilled solution X gms of KOH ( s ) is added, so that molality of OH − ion in the resulting solution becomes 5. Find X ?

TestHub · Accepted Answer

Initial NaOH solution: 75 g of 4% w/w NaOH.

Mass of NaOH = $0.04 \times 75 g = 3 g$ .

Moles of NaOH = $\frac{3 g}{40 g / m o l} = 0.075 m o l$ .

Initial mass of water = $75 g - 3 g = 72 g$ .

After distillation, water collected = 25 g.

Mass of water remaining = $72 g - 25 g = 47 g = 0.047 k g$ .

Moles of OH⁻ from NaOH = 0.075 mol.

Let X be the mass of KOH added. Moles of KOH = $\frac{X}{56 g / m o l}$ .

Total moles of OH⁻ = $0.075 + \frac{X}{56}$ .

Molality of OH⁻ = $\frac{T o t a l m o l eso f O H ^{-}}{M a sso f w a t er ink g} = 5$ .

$5 = \frac{0.075 + \frac{X}{56}}{0.047}$ .

$0.235 = 0.075 + \frac{X}{56}$ .

$0.160 = \frac{X}{56}$ .

$X = 0.160 \times 56 = 8.96 \approx 9 g$ .

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