On dissolving 0.25 g of a non-volatile substance in 30 mL of benzene (density 0.8 g `"mol"^(-1)`) its freezing point decreases by `0.40^(@)C`. Calculate the molecular mass of the non-volatile substance. (`K_(f)` for benzene is 5.12 k `m^(-1)`)

Explanation

We are given that
Mass of non-volatile solute `(w_(B))=0.25g`
Volume of benzene = 30 mL
Density of benzene = 0.8 g `"mol"^(-1)`
Depression in freezing point `(DeltaT_(f))=0.40^(@)C`
Molal freezing constant `(K_(f))=2.15"K kg mol"^(-1)`
We are to find out molecular mass of the non-volatile substance `(M_(B))`
We know that
Mass of solvent `(w_(A))="Density"xx"volume",=0.8"g mol"^(-1)xx30"mL"=24g`
Using the relation, `M_(B)=(K_(f)xxw_(B)xx1000)/(w_(A)xxDeltaT_(f))`
`rArrM_(B)=(5.12xx0.25xx1000)/(24xx0.40)=133.33" g mol"^(-1)`
Therefore, molecular mass of the non-volatile solute is 133.33 g `"mol"^(-1)`.