Chemistry
Acids and Bases

# How many moles of NaCl are present in 300ml of 0.15m solution?

181920 ###### 2012-03-30 01:16:16

Molarity = moles of solute/Liters of solution ( 300 ml = 0.300 Liters )

For our purposes, Moles of solute = Liters of solution * Molarity

Moles NaCl = 0.300 Liters * 0.15 M

= 0.05 moles NaCl

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