LilRu1
LilRu1 LilRu1
  • 19-07-2017
  • Chemistry
contestada

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

Respuesta :

Willchemistry
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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