The boiling point `T'_(b)` of a solvent becomes `T_(b)` on addition of `X_(1)` mole fraction of solute. Heat of vaporisation of solvent is `DeltaH_("vap")`. The relation between elevation in b.pt. `DeltaT_(b)=(T_(b)-T'_(b))` can be given by:
`(DeltaT_(b))/(T_(b)xxT'_(b))=-(2lnX_(1))/(DeltaH_(vap))`
A graph plotted between `log_(10)X_(1) vs (1)/(T_(b))` gives:

Options

A. a straight line with slope `(R )/(DeltaH_(vap))` and intercept `T'_(b)`
B. a stright line with slope `(2.303R)/(DeltaH_(vap))` and intercept `(1)/(T_(b))`
C. a straight line with slope `(DeltaH_(vap))/(DeltaH_(vap))` and intercept `(1)/(T'_(b)xx2.303)`
D. a straight line with slope `(DeltaH_(vap))/(2.303R)` and intercept `-(DeltaH_(vap))/(2.303 RT'_(b))`

Answer

D

Explanation

`(T_(b) -T'_(b))/(T_(b) xx T'_(b)) = -(2.303R log_(10) X_(1))/(Delta H_(vap.))`
`(1)/(T'_(b))-(1)/(T_(b)) = - (2.303 R log_(10) X_(1))/(Delta H_(vap.))`
or `(1)/(T_(b))-(1)/(T'_(b)) = (2.303 R log_(10) X_(1))/(Delta H_(vap.))`
`log_(10)X_(1) = (Delta H_(vap.))/(2.303 RT_(b))-(Delta H_(vap.))/(2.303RT'_(b))`