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Q.Calculate molality of a solution of 5.0 g of ethanoic acid (CH3COOHCH_3COOH) in 150.0 g of benzene.

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2025Subjective· 2mImportance★★★★★
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0.08330.0833 mol of acetic acid in 0.1500.150 kg of benzene gives a molality of about 0.556 m.

Concept. Molality is the number of moles of solute per kilogram of solvent:

m=nsoluteWsolvent (kg).m=\frac{n_{\text{solute}}}{W_{\text{solvent (kg)}}}.

Step 1 — moles of solute. Molar mass of ethanoic acid CH3_3COOH =2(12)+4(1)+2(16)=60=2(12)+4(1)+2(16)=60 g mol−1^{-1}.

n=5.060=0.0833 mol.n=\frac{5.0}{60}=0.0833\ \text{mol}.

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