To prepare 380 mL of 0.500 M NaCl (MW = 58.5 g/mol), how many grams of sodium chloride is required?

This answer did not come from me but here you go!
44 g NaCl
Explanation:
The problem provides you with the molarity and volume of the target solution, so your first step here will be to use this information to figure out how many moles of sodium chloride,
NaCl
, it must contain.
Once you know that, use the compound's molar mass to convert the number of moles to grams.
So, molarity is defined as the number of moles of solute per liter of solution. In your case, a
0.77 M
solution will contain
0.77
moles of sodium chloride, your solute, in
1.0 L
of solution.
Your target solution has a volume of
985
mL
⋅
1 L
10
3
mL
=
0.985 L
which means that it will contain
0.985
L solution
⋅
= 0.77 M

0.77 moles NaCl
1
L solution
=
0.75845 moles NaCl
Sodium chloride has a molar mass of
58.44 g mol
−
1
, which basically means that one mole of sodium chloride has a mass of
58.44 g
.
In your case,
0.75845
moles of sodium chloride will have a mass of
0.75845
moles NaCl
⋅
58.44 g
1
mole NaCl
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
44 g
a
a
∣
∣

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