Question

What is the mole fraction of sodium chloride and water in a solution, prepared by mixing 256 g of sodium chloride and 176 g water? (Na=23g/mol Cl= 35g/mol H=1g/mol o=16g/mol)

Answer 1

$M(NaCl)=23+35=58 g/mol$

$n(NaCl)=256/58=4.4 mol$

$M(H_2O)=2(1)+16=18 g/mol$

$n(H_2O)=176/18=9.8 mol$

$w(NaCl)=n(NaCl)/(n(NaCl)+n(H_2O))=4.4/(4.4+9.8)=0.31$

$w(H_2O)=n(H_2O)/(n(NaCl)+n(H_2O))=9.8/(4.4+9.8)=0.69$