What mass (in g) of NaCl is needed to prepare 1.5 L of a 3.0 M salt water solution?

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Answer 1 · 2015-03-20T01:45:03

263.2g are required.

The number of moles of NaCl in 1.5L of a 3.0M solution #=1.5xx3.0=4.5#

Using:

#A_rNa=23#

#A_rCl=35.5#

#M_rNaCl=23+35.5=58.5#

So 1 mole NaCl weighs 58.5g

So the weight of 4.5 moles = 4.5 x 58.5 = 263.25g

Answer 2 · 2015-03-20T15:29:18

The formula that you will use is:
Grams = Molarity X liters X formula mass of solute

So for your problem:
grams = 3 X 1.5 X 58