Question

Determine the freezing point of a solution containing 0.625 g of glucose (C₆H₁₂O₆) dissolved in 102.8 g of water. 

(Freezing point of water = 273 K, K_f for water = 1.87 K kg mol^-1, at. wt. C = 12, H = 1, O = 16)

Answer 1

M_B = 180 U

ΔT_f = K_f. `("W"_"B" xx 1000)/("M"_"B" xx "W"_"A")`

`= (1.87 xx 0.625 xx 1000)/(180 xx 102.8)`

`= (1.87 xx 625)/(180 xx 102.8) "K" = 0.06 "K"`

T_f' = 273 - 0.06 = 272.94 K