How many moles of solute are in 425ml of 3.0M? How many grams of MgCl2 is this?

Apr 27, 2018

Number of moles $n \approx 1.3$ $\text{mol}$

Amount of mass $m \approx 121.4$ $\text{g}$

Explanation:

The formula relating concentration with number of moles and volume is:

$c = \frac{n}{V}$

$c$ is the concentration, $n$ is the number of moles, and $V$ is the volume (in litres).

Now, we are given values for the volume and concentration, namely $425$ $\text{mL}$ and $3.0$ $\text{M}$.

So let's plug these values into the equation:

$R i g h t a r r o w 3.0 = \frac{n}{425 \cdot {10}^{- 3}} \text{ " }$ (volume given in litres)

$R i g h t a r r o w 3.0 = \frac{n}{0.425}$

Multiplying both sides by $0.425$:

$R i g h t a r r o w 3.0 \cdot 0.425 = \frac{n}{0.425} \cdot 0.425$

$\therefore n = 1.275$ $\text{mol}$

So, there are around $1.3$ moles of solute in $425$ $\text{mL}$ of $3.0$ $\text{M}$.

Then, let's find the mass.

Before we can do this, we need to find the molar mass ${M}_{m}$ of ${\text{MgCl}}_{2}$:

$R i g h t a r r o w {M}_{m} = \left(24.305 + 2 \cdot 35.453\right)$ ${\text{g" * "mol}}^{- 1}$

$R i g h t a r r o w {M}_{m} = 95.211$ ${\text{g" * "mol}}^{- 1}$

Now, the formula that relates number of moles to mass and molar mass is:

$n = \frac{m}{{M}_{m}}$

$n$ is the number of moles, $m$ is the mass, and ${M}_{m}$ is the molar mass.

Plugging values into the equation:

$R i g h t a r r o w 1.275 = \frac{m}{95.211}$

Solving for $m$:

$R i g h t a r r o w 1.275 \cdot 95.211 = \frac{m}{95.211} \cdot 95.211$

$\therefore m = 121.394025$ $\text{g}$

So, the mass is around $121.4$ grams.