Question

Perform the calculations. Round using significant figures. (See companion material attached.)

What is the molarity of a solution containing 0.85 mol of MgCl2 in 1.5 L of solution?

Answer 1

To calculate the molarity (M) of a solution, you can use the formula:

\[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}
\]

In this case, you have 0.85 mol of MgCl₂ and a solution volume of 1.5 L.

Plugging in the values:

\[
\text{M} = \frac{0.85 \, \text{mol}}{1.5 \, \text{L}} = 0.5666667 \, \text{M}
\]

Now, we need to round this value according to the significant figures. The number of significant figures is determined by the values used in the calculation. The value 0.85 has 2 significant figures and 1.5 has 2 significant figures as well. Therefore, we will round our result to 2 significant figures.

Thus, rounding 0.5666667 to 2 significant figures gives:

\[
\text{M} = 0.57 \, \text{M}
\]

So, the molarity of the solution is **0.57 M**.