Question

If the mole fraction of k2so3 in an aqueous solution is 0.0328 what is the weight/weight % of k2so3

Answer 1

Answer: 23

Step-by-step explanation:

Answer 2

`w/w\%=23\%`

Step-by-step 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:

`n_T=n_(K_2SO_3)+n_(H_2O)\\n_(H_2O)=1mol-0.0328mol=0.9672mol`

Thus, one calculates each compound's grams by using their molar masses as:

`m_(K_2SO_3)=0.0328mol*(158.26g)/(1mol)=5.19g\\m_(H_2O)=0.9672mol*(18g)/(1mol)=17.41g`

Therefore, its weight/weight % turns out:

`w/w\%=(5.19g)/(5.19+17.41g)*100 \%=23\%`

Best regards.