Question

How many grams of Na2SO4 are required to make 0.30 L of 0.500 M Na2SO4?

Answer 1

Answer: 21.3 g of $Na_2SO_4$ are required to make 0.30 L of 0.500 M $Na_2SO_4$.

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

moles of $Na_2SO_4$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{142g/mol}$

Now put all the given values in the formula of molality, we get

$0.500M=\frac{xg}{142g/mol\times 0.30L}$

$x=21.3g$

Therefore, 21.3 g of $Na_2SO_4$ are required to make 0.30 L of 0.500 M $Na_2SO_4$.