Question

Calculate the mass of sodium acetate $(C{H_3}COONa)$ required to make $500ml$ of $0.375$ molar aqueous solution. Molar mass of sodium acetate is $82.0245gmo{l^{ - 1}}$ .

Answer 1

Hint: To solve these kinds of questions the most used and best way is the molar approach. It provides us the best option to compare the values directly without any interconversion.

Complete step by step answer: First step is to find the molar concentration:
Moles of the solute in a solution = $molarity$ $\times$ $volume(l)$
$1 litre = 1000 ml$
So $1ml$ = $\dfrac{1}{1000}$ $l$
$0.375$ molar aqueous solution of $(C{H_3}COONa)$ $ = $ $0.375$ moles of $C{H_3}COONa/1000mol$
Now, no. of moles of the $C{H_3}COONa$ in $500ml$ will be:
$
   = \dfrac{{0.375}}{{1000}}*500 \\
   = 0.1875moles \\
 $
Molar mass of $C{H_3}COONa$ is $82.0245gmo{l^{ - 1}}$
Therefore, the mass of the $C{H_3}COONa$ required will be:
$
   = 82.0245*0.1875 \\
   \approx 15.38g \\
 $

Note: Most important thing here is to write the balanced chemical equations as it is the initial and generally most careless state. Apart from that it is necessary to find the correct molar concentration. It is most common mistake of the wrong interconversion of the moles that lead to various wrong answers despite correct execution. Unit conversion should be taken care of as the volume in the $molarity$ is in $litres$ and $ml$ should always be converted into $litres$ and then should be plugged into the formula.