# How many moles of potassium chloride are present in a 3*L volume of 1*mol*L^-1 KCl(aq)?

Jun 3, 2017

Well, I gets a $3 \cdot \text{mole}$ quantity of $K C l$.......

#### Explanation:

$\text{Molarity"="Moles of solute"/"Volume of solution}$.........

And thus for $\text{moles of solute}$, we take the product,

$\text{Molarity"xx"Volume of solution}$

$= 3 \cdot m o l \cdot \cancel{{L}^{-} 1} \times 1000 \cdot \cancel{m L} \times {10}^{-} 3 \cdot \cancel{L} \cdot \cancel{m {L}^{-} 1}$

$= 3.0 \cdot m o l$

Given that $\text{Avogadro's number} \equiv 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$, how many ATOMS are in this quantity?