At What temperature will a sucrose solution boy if it contains 1.55 Mol of sucrose in 600 mL of water? Kb of water is 0.51 °C/m
ΔTb=Kb×m=0.51×1.55×1000600=1.3175ΔTb=T1−T0T1=ΔTb+T0=1.32+100=101.32 °CΔT_b = K_b \times m \\ = 0.51 \times \frac{1.55 \times 1000}{600} = 1.3175 \\ ΔT_b = T_1-T_0 \\ T_1 = ΔT_b + T_0 \\ = 1.32 + 100 = 101.32 \;°CΔTb=Kb×m=0.51×6001.55×1000=1.3175ΔTb=T1−T0T1=ΔTb+T0=1.32+100=101.32°C
Answer: 101.32 °C
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