# How many moles of Ca(NO_3)_2 are there in 75 mL of 0.25 M solution?

Jun 23, 2017

$0.019$ ${\text{mol Ca(NO"_3")}}_{2}$

#### Explanation:

We're asked to find the number of moles of calcium nitrate are in $75$ $\text{mL}$ of a $0.25$ molar solution of ${\text{Ca(NO"_3")}}_{2}$.

To solve this equation, we can use the molarity equation:

$\text{molarity" = "moles solute"/"liters solution}$

We need the solution's volume to be in $\text{L}$, so let's convert it:

75cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = 0.075 $\text{L}$

We're asked to find the number of moles, so

"moles solute" = ("molarity")("liters solution")

"mol Ca(NO"_3")"_2 = (0.25"mol"/(cancel("L")))(0.075cancel("L"))

 = color(red)(0.019 color(red)("mol Ca(NO"_3")"_2

rounded to two sig figs.

Therefore, there are color(red)(0.019 sfcolor(red)("moles of calcium nitrate" in $75$ $\text{mL}$ of a $0.25 M$ solution.