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Problems · Problem 2.6

Q.The normal boiling point of ethyl acetate is 77.06 ⁰C. A solution of 50 g of a nonvolatile solute in 150 g of ethyl acetate boils at 84.27 ⁰C. Evaluate the molar mass of solute if KbK_b for ethyl acetate is 2.77 ⁰C kg mol⁻¹.

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ΔTb=84.27−77.06=7.21\Delta T_b = 84.27 - 77.06 = 7.21 K; M2=1000×2.77×507.21×150=128M_2 = \dfrac{1000 \times 2.77 \times 50}{7.21 \times 150} = 128 g mol⁻¹.

Step 1. Boiling point elevation: ΔTb=84.27−77.06=7.21\Delta T_b = 84.27 - 77.06 = 7.21 ⁰C =7.21= 7.21 K (a temperature DIFFERENCE has the same value in ⁰C and K).

Step 2. From ΔTb=Kbm\Delta T_b = K_b m with m=1000W2M2W1m = \dfrac{1000W_2}{M_2W_1} (Eq. 2.15): M2=1000 KbW2ΔTb W1M_2 = \dfrac{1000\,K_bW_2}{\Delta T_b\,W_1}.

Step 3. Substitute Kb=2.77K_b = 2.77 K kg mol⁻¹, W2=50W_2 = 50 g, W1=150W_1 = 150 g: M2=1000×2.77×507.21×150=1385001081.5=128M_2 = \dfrac{1000 \times 2.77 \times 50}{7.21 \times 150} = \dfrac{138500}{1081.5} = 128 g mol⁻¹.

✓Final answer

Molar mass of the solute M2=128M_2 = 128 g mol⁻¹.

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