Calculate molality of 2.5 g of ethanoic acid (CH3COOH) in 75g of benzene.

Molar mass of C2H4O2:12×2+1×4+16×2=60gmol−1Moles of C2H4O2=2.5g60gmol−1 = 0.0417molMass of banzene in Kg = 75g/1000gkg−1=75×10−3kg Molality of C2H4O2 molesof C2H4O2kg of benzene =0.0417mol×1000gkg−175g=0.556molkg−1

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Calculate molality of 2.5 g of ethanoic acid (CH₃COOH) in 75g of benzene.

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Detailed Solution

Molar mass of $C_{2} H_{4} O_{2} :$
$12 \times 2 + 1 \times 4 + 16 \times 2 = 60 g {mol}^{- 1}$
Moles of $C_{2} H_{4} O_{2} = \frac{2.5 g}{60 g {mol}^{- 1}}$ = 0.0417mol
Mass of banzene in Kg = $75 g / 1000 g {kg}^{- 1}$
$= 75 \times 10^{- 3} kg$
$Molality of C_{2} H_{4} O_{2} \frac{molesof C_{2} H_{4} O_{2}}{kg of benzene}$
$= \frac{0.0417 mol \times 1000 g {kg}^{- 1}}{75 g} = 0.556 mol {kg}^{- 1}$