Find the freezing point of the solution containing 3.6 g of glucose di

1.	Find the freezing point of the solution containing 3.6 g of glucose dissolved in 50 g of H2O. (Kf for H2O = 1.86 K/m).
Answer» Mass of glucose, W_B = 3.6 g Molecular mass of glucose = 180 g mol^-1 Mass of solvent, W_A = 50 g K_f for H₂O = 1.86 K/m = 1.86 K kg^-1 mol^-1 M_B = \(\frac{1000K_f\,\times\,W_B}{W_A\,\times\,\bigtriangleup T_f}\) ∴ \(\bigtriangleup T_f\) = \(\frac{1000K_f\,\times\,W_B}{W_A\,\times\,M_B}\) = \(\frac{1000\,g\,kg^{-1}\,\times\,1.86\,K\,kg^{-1}\,mol^{-1}\,\times\,3.6\,g}{50\,g\times180\,g\,mol^{-1}}\) = 0.744 K. i.e., ΔT_f = T°_f – T_f = 0.744 K ∴ Freezing point of the solution, T_f = T°_f – ΔT = 273 K – 0.744 K = 272.3 K

Find the freezing point of the solution containing 3.6 g of glucose dissolved in 50 g of H2O. (Kf for H2O = 1.86 K/m).

Answer»

Mass of glucose, W_B = 3.6 g

Molecular mass of glucose = 180 g mol^-1

Mass of solvent, W_A = 50 g

K_f for H₂O = 1.86 K/m

= 1.86 K kg^-1 mol^-1

M_B = \(\frac{1000K_f\,\times\,W_B}{W_A\,\times\,\bigtriangleup T_f}\)

∴ \(\bigtriangleup T_f\) = \(\frac{1000K_f\,\times\,W_B}{W_A\,\times\,M_B}\)

= \(\frac{1000\,g\,kg^{-1}\,\times\,1.86\,K\,kg^{-1}\,mol^{-1}\,\times\,3.6\,g}{50\,g\times180\,g\,mol^{-1}}\)

= 0.744 K.

i.e., ΔT_f = T°_f – T_f

= 0.744 K

∴ Freezing point of the solution,

T_f = T°_f – ΔT

= 273 K – 0.744 K

= 272.3 K

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