Answer to Question #318910 in General Chemistry for Adrianna Fernandez

Question

A 30.4ml sample of 2.38 M solution of KCI (molar mass = 74.55 g/mol) is prepared. How many grams of KCI are in the solution?

Answer 1

V(KCl) = 30.4 ml = 0.0304 L

C_M(KCl) = 2.38 M

M(KCl) = 74.55 g/mol

m(KCl) - ?

Answer:

C_M = n/V

n(KCl) = C_M(KCl)*V(KCl)=2.38 M*0.0304 L = 0.072 mol

m(KCl) =n(KCl)*M(KCl)= 0.072 mol*74.55 g/mol = 5.37 g