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,

$\begin{gathered} Volume=\frac{mass}{density} \\ Volume=\frac{100g}{0.998\frac{g}{mL}}=100.2004mL=0.1002L \end{gathered}$

Now, calculating no. of moles

$no.\text{ }of\text{ }moles=\frac{15g}{46.07\frac{g}{mol}}=0.3256mol$

Finally, replacing the values in the formula for Molarity

$Molarity=\frac{0.3256mol}{0.1002L}=3.2495\frac{mol}{L}$

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

$Molality=\frac{0.3256mol}{0.085Kg}=3.8306\text{ }m$

c. Xa (mole fraction)

$no.\text{ }of\text{ }moles\text{ }of\text{ }water=\frac{85g}{18.02\frac{g}{mol}}=4.72mol$

Now, the total moles would be

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

Then, the mole fraction

$\frac{0.3256}{5.0456}=0.0645$

ANSWER:

a. 3.2495 mol/L

b. 3.8306 m

c. 0.0645