Answer to Question #297382 in Chemistry for Leyman

Question

How many grams of KOH are needed in order to prepare 1 litre of a

0.25M solution?

Answer 1

C_M = n / V

n (KOH) = C_M x V = 0.25 x 1 = 0.25 mol

n = m / M

m = n x M

M (KOH) = 56.1 g/mol

m (KOH) = 0.25 x 56.1 = 1.3 g