How many moles of NaCI are present in 0.75L of a 0.55M solution?

,4125 (or .41 if you go by sig figs)

Explanation:

M = (mol)/(L) is a formula for concentration, but we need to solve for moles, so rewrite it as

mol = (L)(M), where we can simply plug in values as follows:

# of mols = (.75L)(.55M) = .4125mol NaCl

The question has 2 significant figures, so you could also put .41 mol NaCl

i think it's c. i'm not sure but i think so.