1.

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) )

Answer»

`0^(@)C `
`- 1.86^(@)C `
`1.86^(@)C `
` - 0.52^(@)C `

Solution :`(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 `


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