Question

A chemist prepares a solution of sodium carbonate (Na2CO3) by measuring out 28 μgol sodium carbonate into a 300ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in μmol of the chemist's sodium carbonate solution. Round your answer to significant digits.

Answer 1

Question in incomplete , complete question is;

A chemist prepares a solution of sodium carbonate by measuring out  28μg of sodium carbonate into a  300 mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in μmol/L of the chemist's sodium carbonate solution. Be sure your answer has the correct number of significant digits.

Answer:

$0.8805 \mu mol/L$ is the concentration of the chemist's sodium carbonate solution.

Explanation:

Mass of sodium carbonate = 28μg =$28\times 10^{-6} g$

$1 \mu g= 1\times 10^{-6} g$

Volume of the solution = 300 mL = 300 × 0.001 L= 0.3 L

( 1 mL = 0.001 L)

Moles of sodium carbonate = $\frac{28\times 10^{-6} g}{106 g/mol}=2.6415\times 10^{-7} mol$

$Concentration = \frac{Moles}{Volume (L)}$

$[Na_2CO_3]=\frac{2.6415\times 10^{-7} mol}{0.3 L}$

$1 mol = 10^6 \mu mol$

$[Na_2CO_3]=\frac{2.6415\times 10^{-7}\times 10^6 \mu mol}{0.3 L}$

$=0.8805 \mu mol/L$

$0.8805 \mu mol/L$ is the concentration of the chemist's sodium carbonate solution.