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The molal elevation constant of water is 0.51 Km–1 The boiling point of 0.1 molal aqueous NaCl solution is nearly :Options1. 100.05°C2. 100.1°C3. 100.2°C4. 101.0°C
Dear student , Use the formula : ΔTb = Kb m Tb – Tbo = 0.51 * 0.1 =0.051 Since water boils at 373.15 K , So Tbo = 373.15 K Therefore , Tb - 373.15 = 0.051 Tb = 373.201 K ( bp of aq. NaCl sol.)Chemistry ConsultantAskiitians
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