Addition of 0.643g of a compound to 43.95g of benzene lowers the freezing point from `5.51^@`C to `5.03^@`C. If `K_f` for benzene is 5.12K kg `mol^-1` , calculate the molar mass of the compound.

Explanation

`W_B = 0.643 g`
`W_A = 43.95 g`
`DeltaT_f = 0.480^@C`
`K_f =5.12` k/m
`M_B = ?`
`DeltaT_f=K_fxxM`
`DeltaT_f=(K_fxxW_Bxx1000)/(W_AxxW_B)`
`M_B = (5.12 xx0.643xx1000)/(43.95xx0.480)`
`=(3292.16)/(21.096)`
`= 156 mu`