(a) Calculate the freezing point of solution when 1.9 g of `MgCl_(2)` (M = 95 g `Mol^(-1)`) was dissolved in 50g of water, assuming `MgCl_(2)` undergoes complete ionization. (`K_(f)` for water = 1.86 K kg `mol^(-1)`).
(b) (i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution?

Explanation

`Delta T_(f) = (ixx K_(f) xx w_(B) xx 1000)/(M_(B) xx w_(B))`
`w_(B) = 1.9 g, w_(A) = 50 g, M_(B) = 95 g mol^(-1),K_(f) = 1.86 Km^(-1)`
`MgCl_(2)` undergoes complete ionization as .
` MgCl_(2) to Mg^(2+) + 2Cl^(-)`
`I = 3`
`therefore " " Delta T_(f) = (3 xx 1.86xx 1.9 xx 1000)/(95 xx 50)`
` = 2.232 K`
Freezing point of solution ` = 273- 2.232 = 270.768 K`