How many moles of potassium chloride are present in a $3 \cdot L$ $3L$ volume of $1 \cdot m o l \cdot L^{- 1}$ $1mol*L^-1$ $K C l (a q)$ $KCl(aq)$ ?

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Answer 1 · 2017-06-03T14:06:48

Well, I gets a $3 \cdot mole$ $3*"mole"$ quantity of $K C l$ $KCl$ .......

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

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

$Molarity \times Volume of solution$ $"Molarity"xx"Volume of solution"$

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

$=3.0*mol$

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

How many moles of potassium chloride are present in a 3*L3⋅L3*L volume of 1*mol*L^-11⋅mol⋅L−11*mol*L^-1 KCl(aq)KCl(aq)KCl(aq)?