To determine the units of the student's answer, let's carefully analyze the expression and cancel out the units step by step:
[tex]\[
\frac{(197 \, \text{g}) \left( \frac{1 \, \text{mL}}{10^{-3} \, \text{L}} \right)}{\left(84.25 \frac{\text{g}}{\text{mol}}\right) \left(5.4 \frac{\text{mol}}{\text{L}} \right)} = ? \, \square
\][/tex]
1. Numerator Analysis:
- The expression inside the numerator is [tex]\( 197 \, \text{g} \times \left( \frac{1 \, \text{mL}}{10^{-3} \, \text{L}} \right) \)[/tex].
- Converting 1 mL to liters: [tex]\( 1 \, \text{mL} = 10^{-3} \, \text{L} \)[/tex].
- Thus, [tex]\( \frac{1 \, \text{mL}}{10^{-3} \, \text{L}} = 10^3 \, \text{L} \)[/tex].
Now the numerator becomes:
[tex]\[
197 \, \text{g} \times 10^3 \, \text{L} = 197 \times 10^3 \, \text{g} \cdot \text{L}
\][/tex]
2. Denominator Analysis:
- The expression inside the denominator is [tex]\( 84.25 \frac{\text{g}}{\text{mol}} \times 5.4 \frac{\text{mol}}{\text{L}} \)[/tex].
- Simplifying the units: [tex]\( \frac{\text{g}}{\text{mol}} \times \frac{\text{mol}}{\text{L}} = \frac{\text{g} \cdot \text{mol}}{\text{mol} \cdot \text{L}} = \frac{\text{g}}{\text{L}} \)[/tex].
Now the denominator becomes:
[tex]\[
84.25 \times 5.4 \frac{\text{g}}{\text{L}} = 454.95 \frac{\text{g}}{\text{L}}
\][/tex]
3. Putting it Together:
Our entire expression now looks like this:
[tex]\[
\frac{197 \times 10^3 \, \text{g} \cdot \text{L}}{454.95 \frac{\text{g}}{\text{L}}}
\][/tex]
4. Canceling the Units:
- The grams (g) in the numerator and denominator cancel out.
- [tex]\( \text{g} \cdot \text{L} \)[/tex] in the numerator divided by [tex]\( \text{g} \cdot \text{L}^{-1} \)[/tex] in the denominator:
[tex]\[
\frac{\text{g} \cdot \text{L}}{\text{g} \cdot \text{L}^{-1}} = \text{L}
\][/tex]
Therefore, the units of the student's answer are [tex]\( \boxed{\text{L}} \)[/tex].
