Question

A chemist makes up a solution by dissolving 39.0 g of Mg(NO3)2 in enough water to produce a final solution volume of 257 mL. To calculate the molarity of the solution, what additional information is needed

Answer 1

Answer:

$M=1.02M$

Explanation:

Hello.

In this case, since the molarity of a solution is computed as the moles of solute over volume of solution:

$M=\frac{n_{solute}}{V}$

Since the volume of the solution is 257 mL or also 0.257 L, we compute the moles of solute, which is magnesium nitrate (molar mass = 148.3 g/mol), as shown below:

$n_{solute}=39.0gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.3 gMg(NO_3)_2}=0.263molMg(NO_3)_2$

Thus, the required molarity turns out:

$M=\frac{0.263molMg(NO_3)_2 }{0.257L}\\ \\M=1.02M$

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