  Question

# 64 g of a mixture of NaCl and KCl were treated with concentrated sulphuric acid. The total mass of the metal sulphates obtained was found to be 76 g. What is the mass percent of NaCl in the mixture? (Molar mass of KCl =74.5gmol−1, NaCl = 58.5 gmol−1, Molar mass of Na2SO4 =142 gmol−1 and Molar mass of K2SO4= 174 gmol−1 NaCl+H2SO4→Na2SO4+HClKCl+H2SO4→K2SO4+HCl  42.85% 84.91%31.50%63.17%

Solution

## The correct option is A 42.85%Consider the mass of NaCl to be x g. Moles of NaCl = x58.5  and moles of KCl = 64−x74.5    Applying the principle of atom conservation (POAC) to Na, Moles of sodium in NaCl = moles of sodium in Na2SO4 1 ×moles of  NaCl = 2 ×moles of Na2SO4 Moles of  Na2SO4 = 12×moles of NaCl Applying the principle of atom conservation (POAC) to K, Moles of potassium in KCl = moles of potassium in K2SO4 1 ×moles of  KCl = 2 × moles of  K2SO4 Moles of K2SO4 = 12×moles of KCl Total weight of Na2SO4 and K2SO4 is 76 g Hence, 12×x58.5×142+12×64−x74.5×174=76    1.214x + 74.738 - 1.168x = 76   0.046x = 1.262  x = 27.43 g  % mass of NaCl =  27.4364×100=42.85%  Suggest corrections   