Question

What is the molarity of the solution prepared by concentrating, by evaporation of solvent, 755 mL of 0.200% (m/v) NaNO₃ to a volume of 629 mL?

Answer 1

Evaporation removes solvent only but not solute. So, amount of solute present in 755 ml remains same in 629ml but percentage changes.

755 ml 0.2% NaNO_​​​3​​ contains 0.2*(755/100) g

= 1.51 g NaNO₃

= (1.51 g/85 g mol^{​​​​-1} )

= 0.017765 mol

[ No. of moles of NaNO_​​​3 = (amount in g)/(molar mass in g/mol)

Molar mass of NaNO₃ = 85 g/mol ]

Molarity = no. Of moles *1000/ volume in mL

= 0.017765 *1000/ 629 mol L^{​​​​​​-1}

  = 0.028 M