What is the mass of #NaCl# required to prepare 0.5 liters of a 2.5 molar solution of #NaCl#?

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Answer 1 · 2016-08-17T20:16:46

73 g

The molecular weight of NaCl is 58.44 g/mol

So, one mole of NaCl weighs 58.44 g.

A 2.5 M solution is 2.5 moles per liter (Molarity is just the number of moles per liter).

Therefore, 0.5 L would contain 1.25 mol. Hence, you would need 1.25 × 58.44 g = 73 g.

As equations:

#M = "moles"/"vol"#

#M = "Molarity"#

#"vol" = "volume (in liters)"#

#"moles" = "grams"/(MW)#

#g = "weight of compound"#

#MW = "molecular weight"#

So,

#M = "moles"/"vol"#

Substitute for moles

#M = (g//MW)/"vol"#

We need #g#, so rearrange

#M × "vol" = g/(MW)#

#g = M × "vol" × MW#

Put in the numbers:

#g = "2.5 mol/L × 0.5 L × 58.44 g/mol"#

#g = "73 g"#

Personally, I prefer the first approach!

You can find out more about moles and molarity here.