Answer to Question #128630 in General Chemistry for Zach Strohmeyer

Question

How many grams of potassium chloride, KCl, are present in 200.0 mL of a 2.50 M solution?

Answer 1

M(KCl) = 74.55 g/mol

Proportion:

2.5 mol – 1000 ml

x mol – 200 ml

x = 0.5 mol

m=n×M

m(KCl) = 0.5 mol × 74.55 g/mol = 37.27 g

Answer: 37.27 g