Answer: 21.3 g of [tex]Na_2SO_4[/tex] are required to make 0.30 L of 0.500 M [tex]Na_2SO_4[/tex].
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n}{V_s}[/tex]
where,
n = moles of solute
[tex]V_s[/tex] = volume of solution in L
moles of [tex]Na_2SO_4[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{142g/mol}[/tex]
Now put all the given values in the formula of molality, we get
[tex]0.500M=\frac{xg}{142g/mol\times 0.30L}[/tex]
[tex]x=21.3g[/tex]
Therefore, 21.3 g of [tex]Na_2SO_4[/tex] are required to make 0.30 L of 0.500 M [tex]Na_2SO_4[/tex].
