Question

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

Answer 1

`\% m/m=10.1\%`

Step-by-step 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*(342g)/(1mol)=112.5g`

`m_(solvent)=1000g`

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

`\% m/m=(112.5g)/(112.5g+1000g)*100\%\\ \\\% m/m=10.1\%`

Regards.