Answer 1

Depression of freezing point and molar mass of the solute:
When W2 gram of solute with molar mass M2 is dissolved in W1 gram of solvent, the molality m is given by

`ΔT_f ∝ m`

`m="Mass of solute in kg"/("Molar mass of solue in kg "×" Mass of solvent in kg")`

`m=W_2/(M_2*W_1)`

∴ The molar mass of a solute using freezing depression method is givenby,

`ΔT_f ∝ m`

`ΔT_f = k_f . ("W"_2)/("M"_2) xx 1000/"W"_1`
Where `K_f` - Cryoscopic constant
`"W"_2` - mass of solute
`"W"_1` - mass of solvent
`"M"_2` - molecular mass of non-volatile solute