Answer
Molality= $0.90\,m$
Mole fraction= $0.016$
Work Step by Step
Mass percent of NaCl= $5.0\%$
$\implies $ 100 g solution contains 5.0 g NaCl.
Mass of water= $100\,g-5.0\,g=95\,g$
Molality= $\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$
$=\frac{\frac{5.0\,g}{58.44\,g/mol}}{95\times10^{-3}\,kg}=0.90\,m$
Mole fraction of NaCl, $X_{NaCl}=\frac{n_{NaCl}}{n_{NaCl}+n_{H_{2}O}}$
$=\frac{\frac{5.0\,g}{58.44\,g/mol}}{\frac{5.0\,g}{58.44\,g/mol}+\frac{95\,g}{18.0\,g/mol}}$
$=0.016$
