Question

a chemist adds 1.05 l of a 0.878 mol/l sodium carbonate (na₂co₃) solution to a reaction flask. calculate the mass in grams of sodium carbonate the chemist has added to the flask. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Calculate moles of sodium carbonate

Use the formula $n = c\times V$, where $n$ is the number of moles, $c$ is the concentration and $V$ is the volume. Given $c = 0.878\ mol/L$ and $V=1.05\ L$.
$n = 0.878\ mol/L\times1.05\ L=0.9219\ mol$

Step2: Calculate molar mass of sodium carbonate

The molar mass of $Na_2CO_3$: $M=(2\times22.99 + 12.01+3\times16.00)\ g/mol=(45.98 + 12.01+48.00)\ g/mol = 105.99\ g/mol$

Step3: Calculate mass of sodium carbonate

Use the formula $m = n\times M$. Substitute $n = 0.9219\ mol$ and $M = 105.99\ g/mol$.
$m=0.9219\ mol\times105.99\ g/mol\approx97.7\ g$ (rounded to 3 significant - digits)

Answer:

$97.7$