Question

What is the molality of a 10% w/v NaCl solution used in the qualitative determination of tannins in plants? (MW NaCl: 58.5 g/mol)

Answer 1

Basis: 100mL NaCl solution

Mass of NaCl in 100 mL solution = 10 g

Mass of solvent = 100 - 10 = 90 g

Moles of solute(NaCl) = $\dfrac{10}{58.5}=0.170\space mol$

Molality = $\dfrac{Moles\space of \space solute}{Mass\space of \space solvent(in\space kg)}=\dfrac{0.170}{90\times10^{-3}}\dfrac{mol}{kg}$

Molality = $1.899\space mol/kg$