How many moles of $N a C l$ $NaCl$ are present in 500 mL of a 1.55 $N a C l$ $NaCl$ solution?

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Answer 1 · 2017-04-13T14:10:05

Molarity= moles of solute/Liters of solution

$1.55 M = \frac{moles}{.5}$ $"1.55 M" = "moles"/".5"$

$moles = 0.775 moles of NaCl$ $"moles = 0.775 moles of NaCl"$

$Molarity = \frac{moles of solute}{Liters of solution}$ $"Molarity" = "moles of solute"/"Liters of solution"$

First, we need to convert $500 mL$ $"500 mL"$ to Liters

$500 ml \cdot \frac{1 L}{1000 ml} = .5 L$ $"500 ml" * "1 L"/"1000 ml" = ".5 L"$

Now we can plug into our formula

$1.55M = \frac{moles}{.5}$ $"1.55M" = "moles"/".5"$

$moles = 0.775 moles of NaCl$ $"moles = 0.775 moles of NaCl"$

Hope it was helpful!

How many moles of NaClNaCl are present in 500 mL of a 1.55 NaClNaCl solution?