How many grams of ch3oh must be added to water to prepare 325 ml of a solution that is 4.50 mch3oh?

Answer is: 46,8 grams of methanol must be added.
V(solution) = 325 ml ÷ 1000 ml/l = 0,325 l.
c(CH₃OH) = 4,5 M = 4,5 mol/L.
n(CH₃OH) = V(solution) · c(CH₃OH).
n(CH₃OH) = 0,325 l · 4,5 mol/l.
n(CH₃OH) = 1,4625 mol.
m(CH₃OH) = n(CH₃OH) · M(CH₃OH).
m(CH₃OH) = 1,4625 mol · 32 g/mol.
m(CH₃OH) = 46,8 g.

False because saturated fatty acids contain only single bond