Question

Ethyl propanoate is an organic molecule that smells like pineapple and has a molar mass of 102.132 g/mol. It is soluble in the organic solvent acetone.

What is the molality of a solution prepared with 91.9 g of ethyl propanoate and 471.0 mL of acetone? The density of acetone is 0.784 g/mL.

Answer 1

The molality of a solution prepared with 91.9 g of ethyl propanoate and 471.0 mL of acetone is 2.44 m.

Step-by-step explanation:

To find the molality of a solution, we use the formula:

molality (m) = moles of solute / kilograms of solvent

First, we need to calculate the mass of the solvent, acetone. Since the density of acetone is 0.784 g/mL, we can calculate the mass of 471.0 mL of acetone as follows:

Mass of acetone = 471.0 mL * 0.784 g/mL = 369.144 g

To convert this mass into kilograms:

Mass of acetone in kg = 369.144 g * (1 kg / 1000 g) = 0.369 kg

Next, we need to calculate the moles of the solute, ethyl propanoate. We know the mass of ethyl propanoate used is 91.9 g and the molar mass is 102.132 g/mol. Hence, the moles of ethyl propanoate are:

Moles of ethyl propanoate = 91.9 g / 102.132 g/mol = 0.90 moles

Finally, we calculate the molality:

Molality of the solution = 0.90 moles / 0.369 kg = 2.44 m (rounded to two decimal places)

Thus, the molality of the solution is 2.44 m.

Answer 2

The molality of the solution prepared with 91.9 g of ethyl propanoate and 471.0 mL of acetone is 2.432 mol/kg.

To find the molality m of the solution, you can use the following formula:

`\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \]`

First, calculate the moles of ethyl propanoate
`(\(C_5H_(10)O_2\))` using its molar mass:

`\[ \text{Moles of ethyl propanoate} = \frac{\text{mass}}{\text{molar mass}} \]`

`\[ \text{Moles of ethyl propanoate} = \frac{91.9 \, \text{g}}{102.132 \, \text{g/mol}} \]` = 0.8991 mol

Now, calculate the mass of acetone used in the solution. Since the density
`(\(\rho\))` of acetone is given as
`\(0.784 \, \text{g/mL}\)`, you can use the formula:

`\[ \text{Mass of acetone} = \text{volume} * \text{density} \]`

`\[ \text{Mass of acetone} = 471.0 \, \text{mL} * 0.784 \, \text{g/mL} \]` = 369.984 g = 0.369984 kg

Now, calculate the molality using the formula mentioned at the beginning:

`\[ m = \frac{\text{moles of ethyl propanoate}}{\text{mass of acetone (in kg)}} \]`

= 0.8991/0.369984

= 2.432 mol/kg

Therefore, the molality of the solution prepared with 91.9 g of ethyl propanoate and 471.0 mL of acetone is approximately 2.432 mol/kg.