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

1 Answer
Jun 3, 2017

Well, I gets a 3*"mole" quantity of KCl.......

Explanation:

"Molarity"="Moles of solute"/"Volume of solution".........

And thus for "moles of solute", we take the product,

"Molarity"xx"Volume of solution"

=3*mol*cancel(L^-1)xx1000*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)

=3.0*mol

Given that "Avogadro's number"-=6.022xx10^23*mol^-1, how many ATOMS are in this quantity?