Question

A solution contains a compound with a molar mass of, 87.82 g/mol. How many gramsof this compound are contained if the solution has a volume of 1,034.9 mL and amolarity of 1.4 M?

Answer 1

The question requires us to calculate the mass of a compound, in grams, contained in 1,034.9 mL of a 1.4 M solution of this compound, given its molar mass corresponds to 87.82 g/mol.

The molarity of a solution can be calculated as (Equation 1):

`C=(n)/(V)`

where C corresponds to the molarity (in mol/L), n corresponds to the number of moles (in mol) and V is the volume of the solution (in L).

Also, the number of moles of a compound can be defined as (Equation 2):

`n=(m)/(M)`

where n is the number of moles (in mol), m is the mass of the compound (in g) and M corresponds to the molar mass of this compound (in g/mol).

We can combining the equations 1 and 2 above and arrive at the following expression:

`C=(m)/(M* V)`

where we are able to calculate the molarity of a solution from its mass (in grams), molar mass (in grams per mol) and volume (in liters).

We can also rearrange this equation in order to obtain the mass of the compound:

`C=(m)/(M* V)\to m=C* M* V`

Note that, in the last equation, we must use the volume of the solution in units of liter (L). Thus, to apply the values given by the question to this equation, we need to convert the volume from mL to L:

1000 mL = 1 L

1,034.9 mL = 1.0349 L

Now, we can use the values of molarity, molar mass and volume from the question to calculate the mass required:

`\begin{gathered} m=C* M* V \\ m=(1.4mol/L)*(87.82g/mol)*(1.0349L) \\ m=127.2g \end{gathered}`

Therefore, the mass of compound contained in the solution given is 127.2g.