Question

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

Answer 1

Number of moles `n approx 1.3` `"mol"`

Amount of mass `m approx 121.4` `"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` `"mL"` and `3.0` `"M"`.

So let's plug these values into the equation:

`Rightarrow 3.0 = frac(n)(425 * 10^(- 3)) " " "` (volume given in litres)

`Rightarrow 3.0 = frac(n)(0.425)`

Multiplying both sides by `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.