Question

A student determines the molar mass of acetone, CH3COCH3, by the method used in this experiment. He found that the equilibrium temperature of a mixture of ice and water was 1.2oC on his thermometer. When he added 12.2 g of her sample to the mixture, the temperature, after thorough stirring, fell to –3.0oC. He then poured off the solution through a screen into a beaker. The mass of the solution was 100.5 g. How much water was in his decanted solution? 

Answer 1

mass of acetone added= 12.2 g

mass of solution =100.5 g, moles of acetone=$\frac{12.2}{58}=0.2103$

change inn temp= 4.2

we know that as per depression of freezing point = 1.86

we know that

$\Delta T_f= K_fm$

4.2= 1.86$\times molality$

$\frac{moles\times 1000}{mass of water}= \frac{4,2}{1.86}$

mass of water=$\frac{0.2103\times 1.86\times 1000}{4.2}$ = 93.15

so mass of water was= 93.15 g