A solution containing 0.5126 g of napththalene `("molar mass"= 128 g mol^(-1))` in 50.0 g of carbon tetrachloride gave a boiling point elecantion of 0.402 K. Find the molar mass of the unknown solute.

Explanation

IN case of naphthalene
Mass of naphthlene `(W_(B))`=0.5126 g
Mass of carbon tetrachloride `(W_(A))`= 50.0g=0.050kg
Molar mass of naphthalene `(M_(B))`=128 g `mol^(-1)`
Elevation in boiling point `(DeltaT_(b))=0.647 K`
`K_(b)=(M_(B)xxDeltaT_(b)xxW_(A))/W_(B)=((128 g mol^(-1))xx(0.402 K)xx(0.50 kg))/((0.5126g))=5.019 K kg mol^(-1)`
In case unknown solute
Upon comparing the available infrmation both `W_(A)andK_(b)` values in two cases are the same. Therefore. by comparing in terms of these values,
`(M_(B)xxDeltaT_(b))/W_(B)("Naphthalene")=(M_(B)xxDeltaT_(b))/W_(B)("Unknown sloute")`
`((128 g mol^(-1)xx(0.402 K))xx(0.402K))/((0.5126 g))=(M_(B)xx(0.647K))/((0.6216 g))`
`M_(B)=((128 g mol^(-1))xx(0.402K)xx(0.6216 g))/((0.5126 g)xx(0.099 kg))=96.44 g mol^(-1)`