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

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Answer 1 · 2018-04-27T10:08:36

Number of moles $n \approx 1.3$ $n approx 1.3$ $mol$ $"mol"$

Amount of mass $m \approx 121.4$ $m approx 121.4$ $g$ $"g"$

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

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

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

Now, we are given values for the volume and concentration, namely $425$ $425$ $mL$ $"mL"$ and $3.0$ $3.0$ $M$ $"M"$ .

So let's plug these values into the equation:

$\Rightarrow 3.0 = \frac{n}{425 \cdot 10^{- 3}}$ $Rightarrow 3.0 = frac(n)(425 * 10^(- 3)) " " "$ (volume given in litres)

$\Rightarrow 3.0 = \frac{n}{0.425}$ $Rightarrow 3.0 = frac(n)(0.425)$

Multiplying both sides by $0.425$ $0.425$ :

$\Rightarrow 3.0 \cdot 0.425 = \frac{n}{0.425} \cdot 0.425$ $Rightarrow 3.0 * 0.425 = frac(n)(0.425) * 0.425$

$therefore n = 1.275$ $"mol"$

So, there are around $1.3$ moles of solute in $425$ $"mL"$ of $3.0$ $"M"$ .

Then, let's find the mass.

Before we can do this, we need to find the molar mass $M_(m)$ of $"MgCl"_(2)$ :

$Rightarrow M_(m) = (24.305 + 2 * 35.453)$ $"g" * "mol"^(- 1)$

$Rightarrow M_(m) = 95.211$ $"g" * "mol"^(- 1)$

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:

$Rightarrow 1.275 = frac(m)(95.211)$

Solving for $m$ :

$Rightarrow 1.275 * 95.211 = frac(m)(95.211) * 95.211$

$therefore m = 121.394025$ $"g"$

So, the mass is around $121.4$ grams.