How many mL of 0.300 M NaF would be required to make a 0.0355 M solution of NaF when diluted to 250.0 mL with water?

Answer 1

(x ml)(0.300 M) = (250 ml)(0.0355 M)

x = 29.58 ml = 29.6 ml (3 significant figures)

Answer 2

The number of moles of the solution before dilution = The number of moles of the solution after dilution

Therefore,

(M*V)_{Before Dil.} = (M*V)_{After Dil.}

(0.3*V₁) = (0.0355*250)

V₁ = 29.6 mL

Answer 3

For a Dilution problem we can use the following formula:

M1xV1 = M2xV2

M1 = 0.0355 Moles NaF

V1 = 250 ML

M2 = 0.0300 Moles NaF

V2=?

(0.0355 M) x ( 250 ML) = (0.0300 M) x V2

Divide both sides by 0.300 and calculate, the moles cancel out to leave you with ML

(0.0355)(250 ML) / 0.300 = 8.875 /0.300

= 29.583 ML