# A solution of 6.42 g of a carbohydrate in 105.5 g of water has a density of 1.024 g/mL and an osmotic pressure of 4.61 atm at 20.0°C. What is the molar mass of the carbohydrate?

Jan 1, 2018

This is a fairly challenging solution problem. I think the only way to solve this is to make the approximation I did. Definitely made me scratch my head!

Let's first find the molarity,

$4.61 a t m = 1 \cdot M \cdot \frac{0.08206 L \cdot a t m}{m o l \cdot K} \cdot 293 K$
$\therefore M \approx 0.192 M$

Then, we'll calculate the percent mass of the carbohydrate,

(6.42g)/(6.42g + 105.5g) approx 5.74%

and assume we have one liter of solution, extrapolate the theoretical mass of the carbohydrate, and go from there,

$\frac{1.024 g}{m L} \cdot {10}^{3} m L = 1024 g \cdot 0.0574 \approx 58.7 g$

of carbohydrate are theoretically in one liter.

Now, we know the molarity, so let's find the molar mass,

$0.192 m o l = 58.7 g \cdot \frac{1}{\text{MM}}$
$\therefore \text{MM} \approx \frac{306 g}{m o l}$

This is a very reasonable and I'm curious to see if anyone has any other ideas. Good question!