Question

Determine the Molarity of a solution made by dissolving 21.6 grams of sodium carbonate, Na₂CO_3, in enough water to make 400.0 ml of solution.

Answer 1

M (Na₂CO₃) = 106 g/mol

m (Na₂CO₃) = 21,6 g

V = 400 ml = 0,4 liters

C_M = ?

1.

$n=\cfrac {m}{M}= \cfrac {21,6 g}{106g/mol} = 0,2mol$

2.

$C_M= \cfrac {n}{V} = \cfrac{0,2mol}{0,4liters}= 0,5mol/l$

Answer: 0,5 mol/l