Answer : The molarity of a solution is, 1.84 M
Explanation : Given,
Mass of \(Na_2SO_4\) = 85.0 g
Molar mass of \(Na_2SO_4\) = 142 g/mole
Volume of solution = 325 ml
Molarity : It is defined as the number of moles of solute present in one liter of solution.
In this question, the solute is \(Na_2SO_4\).
Formula used :
\(Molarity=\frac{w_b}{M_b\times V}\times 1000\)
where,
\(w_b\) = mass of solute \(Na_2SO_4\)
\(M_b\) = molar mass of solute \(Na_2SO_4\)
\(V\) = volume of solution in ml
Now put all the given values in the above formula, we get the molarity of the solution.
\(Molarity=\frac{85.0g}{142g/mole\times 325ml}\times 1000=1.84mole/L=1.84M\)
Therefore, the molarity of the solution is, 1.84 M
