Question

Acetone has a density of 0.7857 g/cm3. What would be the volume in mL of acetone if the mass was 0.3201 lbs? Remember 1 lb = 0.453 kg

A. 184.8 mL
B. 0.8982 mL
C. 251.5 mL
D. 0.1848 mL
E. 114.3 mL

Answer 1

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To solve this problem, we can use the given density of acetone and convert the mass from pounds to grams.

First, let's convert the mass from pounds to kilograms:

`\displaystyle \text{Mass (kg)} = 0.3201 * 0.453 = 0.14517 \,\text{kg}`

Now, we can use the density formula to find the volume in cubic centimeters (cm³):

`\displaystyle \text{Density} = \frac{{\text{Mass}}}{{\text{Volume}}}`

Rearranging the formula to solve for volume, we have:

`\displaystyle \text{Volume (cm}^(3)\text{)} = \frac{{\text{Mass}}}{{\text{Density}}}`

Substituting the given values, we get:

`\displaystyle \text{Volume (cm}^(3)\text{)} = \frac{{0.14517\,\text{kg}}}{{0.7857\,\text{g/cm}^(3)}}`

Now, let's convert the volume from cubic centimeters to milliliters:

`\displaystyle \text{Volume (mL)} = \text{Volume (cm}^(3)\text{)}`

Substituting the value we obtained earlier, we have:

`\displaystyle \text{Volume (mL)} = \frac{{0.14517\,\text{kg}}}{{0.7857\,\text{g/cm}^(3)}}`

Calculating the value, we find:

`\displaystyle \text{Volume (mL)} = 184.77\,\text{mL}`

Therefore, the volume of acetone would be approximately 184.8 mL.

The correct option is A. 184.8 mL.

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