19-07-2017
Chemistry

contestada

How many moles of NaCl are required to prepare 0.80 L of 6.4 M NaCl

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Willchemistry Willchemistry

19-07-2017

Hey there!

Solution :

Volume of solution = 0.80 L

Molarity of solution = 6.4 M

Therefore :

n = M * V

n = 6.4 * 0.80

n = 5.12 moles of NaCl

hope this helps!

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