The freezing point of `1` molal `NaCl` solution assuming `NaCl` to be `100%` dissociated in water is:

Options

A. `-1.86^(@)C`
B. `-3.72^(@)C`
C. `+1.86^(@)C`
D. `+3.72^(@) C `

Answer

B

Explanation

`T_(f) = K_(f) xx m xx i `
`Delta T_(f) = 1.86 xx 1 xx 2 = 3.72,`
`(NaCl = Na^(+) + Cl^(-)` no of particles =2=i)
`Delta T_(f) = T_(f)^(@) - T_(f)`
`:. T_(f) = T_(f)^(@) - Delta T_(f) = 0 - 3.372 = -3.72^(@)C`