Question

type the correct answer in the box. sodium carbonate (na₂co₃) is sometimes used as a water - softening agent. suppose that a worker prepares a 0.724 m solution of na₂co₃ and water. the volume of the solution is 1.421 liters. what is the mass of solute in the solution? express your answer to three significant figures. the mass of dissolved sodium carbonate is grams

Explanation:

Step1: Calculate the number of moles of $Na_2CO_3$

Use the formula $n = M\times V$, where $M$ is molarity and $V$ is volume. Given $M = 0.724\ M$ and $V=1.421\ L$. So $n = 0.724\times1.421$.
$n=0.724\times1.421 = 1.030704\ mol$

Step2: Calculate the molar - mass of $Na_2CO_3$

The molar - mass of $Na$ is $22.99\ g/mol$, $C$ is $12.01\ g/mol$, and $O$ is $16.00\ g/mol$. For $Na_2CO_3$, $M_{Na_2CO_3}=2\times22.99 + 12.01+3\times16.00$.
$M_{Na_2CO_3}=45.98 + 12.01+48.00=105.99\ g/mol$

Step3: Calculate the mass of $Na_2CO_3$

Use the formula $m = n\times M$. Substitute $n = 1.030704\ mol$ and $M = 105.99\ g/mol$. So $m=1.030704\times105.99$.
$m = 1.030704\times105.99=109.244217\ g\approx109\ g$

Answer:

109