Question

PLZ HELP ASAP

Approximately how many grams of potassium chloride(KCI) are required to prepare 600. mL of a 2.50 M solution ?

A) 103 g
B) 112 g
D) 125 g

Answer 1

(B)

Step-by-step explanation:

2.50 M KCl = (2.50 mol KCl/L)(0.600 L) = 1.50 mol KCl

molar mass of KCl = 74.5513 g KCl/mol

1.50 mol KCl × (74.5513 g KCl/mol) = 112 g KCl