Question

what volume of 1.50 M Na2CO3 solution can be prepared from 2.00g of solid Na2CO3?

Answer 1

What volume of 1.50 M Na₂CO₃ solution can be prepared from 2.00 g of solid Na₂CO₃?

Solution: As you know, molar concentration of solution can be calculated as:

$$C(Na_2CO_3) = \frac{m(Na_2CO_3)}{M(Na_2CO_3) \cdot V}$$

, where $C$ is the concentration, mol/L; $m$ is the mass of dissolved sodium carbonate, g; $M$ is the molar mass of sodium carbonate, $M(Na_2CO_3) = 23.2 + 12 + 16.3 = 106\ \mathrm{g/mol}$; $V$ is the volume of solution, L;

Then, $V = \frac{m(Na_2CO_3)}{M(Na_2CO_3) \cdot C(Na_2CO_3)} = \frac{2}{106 \cdot 1.5} = 0.0126\ \mathrm{L} = 12.6\ \mathrm{mL}$

Answer: 12,6 mL.