Answer:
[tex]\% m/m=10.1\%[/tex]
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:
[tex]m_{sucrose}=0.329mol*\frac{342g}{1mol}=112.5g[/tex]
[tex]m_{solvent}=1000g[/tex]
In such a way, we proceed to the calculation of the mass percent as follows:
[tex]\% m/m=\frac{112.5g}{112.5g+1000g}*100\%\\ \\\% m/m=10.1\%[/tex]
Regards.
