How do you calculate the molarity of each solution 1.0 mol KCI in 750 mL of solution and then 0.50 mol MgCl_2 in 1.5 L of solution?

May 13, 2017

With ease.........we use the quotient...

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

Explanation:

And thus $\left[K C l\right] = \frac{1.0 \cdot m o l}{0.750 \cdot L} = 1.33 \cdot m o l \cdot {L}^{-} 1$.

And $\left[M g C {l}_{2}\right] = \frac{0.50 \cdot m o l}{1.50 \cdot L} = 0.33 \cdot m o l \cdot {L}^{-} 1$.

I include the units because when you have to calculate a volume or the moles of solute present, it is fairly easy to decide when to divide and when to multiply, simply by reference to the given units.

What are the concentrations of chloride ions in each solution?