Question

3. A 13.0 % solution of K2CO3 by mass has a density of 1.09 g/cm³ . Calculate the molarity of the solution.

Answer 1

Let the solution be of mass 100 g

So the amount of K₂CO₃ will be 13% of 100 g i.e. 13 g

Molar mass of K₂CO₃ = 138.205 g /mol

n(K₂CO₃) $= \frac{13 }{ 138.205} = 0.094 \; moles$

$Density = \frac{mass }{ volume}$

Volume of the solution $= \frac{mass }{ density}$

$= \frac{100 }{1.09} \\ = 91.74 \; mL \\ Molarity = \frac{no. \;of \;moles }{ volume \;of \;solution\; in \;L} \\ M = \frac{0.094 }{ 0.09174} \\ M = 1.024$

The molarity of the solution is 1.024 M