Question

A solution that contains 55.0 g of ascorbic acid (Vitamin C) in 250. g of water freezes at –2.34°C. Calculate the molar mass (in units of g/mol) of the solute. K f of water is 1.86°C/m

Answer 1

Answer:

$M=174.87g/mol$

Explanation:

Hello,

In this case, we consider the depression in the freezing point as:

$\Delta T=-i*m*Kf$

Whereas the van't hoff factor for ascorbic acid is 1 since it is covalent, thus, we solve for the molality as shown below:

$(-2.34-0)\°C=-m*1.86\°C/m\\\\m=\frac{-2.34\°C}{-1.86\°C/m} =1.26m$

Next, since molal units are mol/kg, we can compute the present moles of ascorbic acid in the 250 g (0.25kg) of water:

$n=1.26mol/kg*0.25kg=0.315mol$

Finally, the molar mass with the given 55.0 g of ascorbic acid:

$M=\frac{55.0g}{0.315mol}\\ \\M=174.87g/mol$

Regards.