Answer:
[tex]w/w\%=23\%[/tex]
Explanation:
Hello,
In this case, since potassium sulfite is in an aqueous solution, that is water as solvent, one could assume 0.0328 moles of potassium sulfite and 1 mole of solution (total), therefore, the moles of water are:
[tex]n_T=n_{K_2SO_3}+n_{H_2O}\\n_{H_2O}=1mol-0.0328mol=0.9672mol[/tex]
Thus, one calculates each compound's grams by using their molar masses as:
[tex]m_{K_2SO_3}=0.0328mol*\frac{158.26g}{1mol}=5.19g\\m_{H_2O}=0.9672mol*\frac{18g}{1mol}=17.41g[/tex]
Therefore, its weight/weight % turns out:
[tex]w/w\%=\frac{5.19g}{5.19+17.41g}*100 \%=23\%[/tex]
Best regards.
