(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

(a) Molar Mass of `MgCl_(2)` = `95g//mol`. (MB)
`W_(B) = 1.9g`
`W_(A) = 50g`,
`K_(f) = 1.86K kg//mol`
i=3
`DeltaT_(f)=iK_(f)xxm`
`overset(@)(T)_(A) - T_(S) = (KfxxW_(B))/(M_(B)xxW_(A)(Kg))xxl`
`O-T_(s) = (1.86 xx 1.9 xx 1000)/(95 xx 50) xx 3`
`-T_(S) = 2.232`
`T_(S) = -2.232 .^(@)C`
(b) (i) 2M glucose has a higher boiling point than 1M glucose because 2M glucose has 2 moles of glucose, as no, of moles increases, boiling increases.
(ii) When external pressure applied, becomes more than the osmotic pressure of solution the solvent will flow from the solution into the pure solvent through semi-permeable membrane. Which is called reverse osmosis. Which is used in the desalination of sea water.