A 5% solution (w/W) of cane sugar (molar mass = 342 g `mol^(-1)` ) has freezing point 271 K. What will be the freezing point of 5% glucose (molar mass = 18 g `mol^(-1)`) in water if freezing point of pure water is 273.15 K ?

Options

A. 273.07 K
B. 269.07 K
C. 273.15 K
D. 260.09 K

Answer

B

Explanation

`DeltaT_f=(K_fxxW_B)/(M_BxxW_A)`
For cane sugar solution, 2.15 K = `(K_fxx5)/(342xx0.095)` ( `because` 95 g of water = 0.095 kg )
For glucose soution, `DeltaT_f=(K_fxx5)/(180xx0.095)`
`DeltaT_f/2.15=(K_fxx5)/(180xx0.095)xx(342xx0.095)/(K_fxx5)`
`DeltaT_f=342/180xx2.15`=4.085 K
Freezing point of glucose solution = 273.15-4.085 = 269.07 K