Answer to Question #194604 in General Chemistry for CAMRYN

Question

What is the molarity of a solution that has 0.75L of 0.25M Na2SO4

Answer 1

Proportion:

0.25 mol – 1.00 L

x mol – 0.75 L

$x = \frac{0.25 \times 0.75}{1} = 0.1875 \;mol$

M(Na₂SO₄) = 142.04 g/mol

$m = n \times M = 0.1875 \times 142.04 = 26.63 \;g$

We need 26.63 g of Na₂SO₄ to prepare 0.75 L of 0.25 M Na₂SO₄.