Answer to Question #244303 in Inorganic Chemistry for swda

Question

5. What is the molality of a 10% by mass NaCl solution when used in qualitative determination of tannins in plants? *

Answer 1

10% means 100 g of solution contains 10 g of NaCl

M = 58.44

n = 10/58.44 = 0.1711 mol

Density of the solution is 1.09

V = 100 / 1.09 = 91.74 mL = 0.09174 L

Molarity = 0.1711 / 0.09174 = 1.86 M

Molality = Weight of solute/molar weight of solute ×mass of solvent in Kg

10/58×0.09174

= 1.88