Question

An aqueous solution is 15.0% methanol (CH;O) by mass and has a density of 0.998 g/mL. Calculate thea. Molarityb. Molalityc. Xa

Answer 1

INFORMATION:

We know that:

- An aqueous solution is 15.0% methanol (CH4O) by mass and has a density of 0.998 g/mL

And we must calculate molarity, molality and Xa

STEP BY STEP EXPLANATION:

First, we must analyze the given information:

15.0% methanol (CH4O) by mass means,

mass of methanol = 15 g

mass of water (solvent) = 85 g = 0.085 kg

mass of solution = 100 g

Now, we can calculate:

a. Molarity

To calculate it, we must use the following formula

`Molarity=\frac{no.\text{ }moles}{volume\text{ }of\text{ }solution\text{ }in\text{ }liters}`

The solution has a density of 0.998 g/mL

Using that,

`\begin{gathered} Volume=(mass)/(density) \\ Volume=(100g)/(0.998(g)/(mL))=100.2004mL=0.1002L \end{gathered}`

Now, calculating no. of moles

`no.\text{ }of\text{ }moles=(15g)/(46.07(g)/(mol))=0.3256mol`

Finally, replacing the values in the formula for Molarity

`Molarity=(0.3256mol)/(0.1002L)=3.2495(mol)/(L)`

b. Molality

To calculate it, we must use the following formula

`Molality=\frac{no.\text{ }of\text{ }moles}{mass\text{ }of\text{ }solvent\text{ }in\text{ }Kg}`

Now, replacing the values in the formula

`Molality=(0.3256mol)/(0.085Kg)=3.8306\text{ }m`

c. Xa (mole fraction)

`no.\text{ }of\text{ }moles\text{ }of\text{ }water=(85g)/(18.02(g)/(mol))=4.72mol`

Now, the total moles would be

`\text{ Total moles}=(4.72+0.3256)mol=5.0456mol`

Then, the mole fraction

`(0.3256)/(5.0456)=0.0645`

ANSWER:

a. 3.2495 mol/L

b. 3.8306 m

c. 0.0645