The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to354.11 K. Calculate the molar mass of the solute. (K_(b) for benzene is 2.53 K kg mol^(-1))

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatil

1.	The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to354.11 K. Calculate the molar mass of the solute. (K_(b) for benzene is 2.53 K kg mol^(-1))
Answer» Solution :Given values are: `T_(("benzene"))^(@) = 353.00 K , K_(b) = 2.53.00 K kg mol_(-1)` `T_(b("solution")) = 354.00 K` `W_("SOLUTE") = 1.80 g` `W_("solvent") = 90 g` The elevation in boiling point ,`DeltaT_(b) = T_(b("solution")) - T_(b("solvent"))^(@)` `= 354.11 - 353.23` `=0.88 K` Molar mass of solute is given as `Mw_("solute") = (K_(b) xx 1000 xx W_("solute"))/(DeltaT_(b) xx W_("solvent")` `Mw_("solute") = (2.53 xx 1000 xx 1.80)/(0.88 xx 90) = 58.0 g mol^(-1)` Hence, the molar mass of solute is `58.0 g mol^(-1)`.