Question

97.30g of a mystery compound is added to 500g of water, raising its boiling point to 100.78°C. What is the molar mass of the mystery compound? kb of water = 0.51°C/m

Answer 1

$\Delta T_b = 100.78^ \circ-100^\circ = 0.78^\circ$

We know,

$m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.78^\circ}{0.512\dfrac{^\circ C}{m}} = 1.5m$

Next, take this molality value and multiply it by given mass of solvent we get 0.76 mole solute.

last, divide the number of grams of the mystery solute by the number of moles, giving you the molecular mass of the compound.

$= \dfrac{97.30}{0.76} = 130g/mol$