Question

What is the mass percent of sucrose (C12H22O11, Mm = 342 g/mol) in a 0.329-m sucrose solution?

Answer 1

Answer:

$\% m/m=10.1\%$

Explanation:

Hello,

In this case given the molal solution of sucrose, we can assume there are 0.329 moles of sucrose in 1 kg of solvent, thus, computing both the mass of sucrose and solvent in grams, we obtain:

$m_{sucrose}=0.329mol*\frac{342g}{1mol}=112.5g$

$m_{solvent}=1000g$

In such a way, we proceed to the calculation of the mass percent as follows:

$\% m/m=\frac{112.5g}{112.5g+1000g}*100\%\\ \\\% m/m=10.1\%$

Regards.