2g of benzoic acid dissolved in 25g of benzene produces a freezing-point depression of 1.62^(@). Calculate the molecular weight. Compare this with the molecular weight obtained from the formula for benzoic acid, C_(6)H_(5)COOH. (K_(f)=4.90)

2g of benzoic acid dissolved in 25g of benzene produces a freezing-poi

1.	2g of benzoic acid dissolved in 25g of benzene produces a freezing-point depression of 1.62^(@). Calculate the molecular weight. Compare this with the molecular weight obtained from the formula for benzoic acid, C_(6)H_(5)COOH. (K_(f)=4.90)
Answer» Solution :Molality `=(DeltaT_(f))/(K_(f))` ……….(Eqn. 7) `=(1.62)/(4.9)` Suppose the molecular weight of benzoic acid is M `:.` moles of benzoic acid `=(2)/(M)` `:.` molality (moles /1000g)`=(2)/(M)XX(1000)/(25)=(80)/(M)` Thus `(80)/(M)=(1.62)/(4.9)`, `M=241.97` The actual value of the molecular weight of benzoic acid, from its formula `C_(6)H_(5)COOH`, is 122. The observed (experimental) value i.e., `241.97` is much greater than the NORMAL value because of the association of benzoic acid in benzene