Question

How many g of K2CrO4 (MM = 194.2 g / mol) are required to prepare 300 mL of a solution with a concentration of 1.50 M of this same compound?

Answer 1

Proportion:

1.50 mol – 1000 mL

x mol – 300 mL

$x = \frac{1.50 \times 300}{1000}= 0.45 \;mol \\ m = 0.45 \times 194.2 = 87.39 \;g$

Answer: 87.39 g