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?

1 answer

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.