The measured freezing point depression of a non-volatile solute in aqueous solution is `0.20^(@) C` . The elevation in boiling point of the same solution will be [`K_(f)` = 1.86 K/m, `K_(b)` = 0.52 K/m]

Options

A. 0.0186
B. 0.056
C. 0.052
D. 5.2

Answer

B

Explanation

`DeltaT_(f) = K_(f) xx m`
` T_f`, freezing point of solvent , m -molality
` Delta T_(b) = K_(b) xx m`
`K_(f) - ` molal depression constant
`(DeltaT_(f))/(DeltaT_(b))=(K_(f))/(K_(b)) implies (0.20)/(DeltaT_(b)) = (1.86)/(0.52)`
` implies DeltaT_(b) = (0.20 xx 0.52)/(1.86) = (104)/(1860) = 0.056`