An aqueous solution of a non-electrolye boils at `100.52^(@)C ` . The freezing point of the solution will be
(`K_(b) = 0.52 K kg mol^(-1), K k_(g) = 1.86 K kg mol^(-1)` )

Options

A. `0^(@)C `
B. `- 1.86^(@)C `
C. `1.86^(@)C `
D. ` - 0.52^(@)C `

Answer

B

Explanation

`(Delta T_(b))/(K_(b)) =( Delta T_(f))/(K_(f))`
`Delta T_(b) = 100.52 - 100 `( B.P. of `H_(2)O = 100^(@)C ` )
`= ( 0.52)/( 0.52) =( DeltaT_(f))/(1.86)`
`DeltaT_(f)= 1.86 ` ( Freezing point of `H_(2)O = 0^(@) C ) `
Freezing point of solution `= - 1.86^(@) C `