Question

1.39

a 6.0 % (m/v) KOH solution prepared from 10.0 mL of a 20.0 % KOH solution

a 1.3 M HCl solution prepared from 25 mL of a 6.0 M HCl solution

a 2.0 M NaOH solution prepared from 50.0 mL of a 12 M NaOH solution

Answer 1

We need to apply dilution law, which is based on the mass conservation principle

initial mass = final mass

this apply for moles as weel ( if there is no reaction, which is the case )

mol of A initially = mol of A finally

or, for this case

moles of A in stock = moles of A in diluted solution

Recall that

mol of A = Molarity of A * Volume of A

then

moles of A in stock = moles of A in diluted solution

Molarity of A in stock * Volume of A in stock = Molarity of A in diluted solution* Volume of A in diluted solution

Now, substitute known data

a)

C1V1 = C2V2

10*20 = 6*V

V = 10*20/6

V = 33.3 mL

b)

C1V1 =C2V2

25*6 = 1.3*V

V = 25*6/1.3

V = 115.4 mL

c)

M1V1 = M2V2

12*50 = 2*V

V = 12*50/2

V = 300 mL