To solve the expression [tex]\(\frac{\left(1.26 \frac{ g }{ mL }\right)\left(\frac{1 mL }{10^{-3} L }\right)}{\left(44.49 \frac{ g }{ mol }\right)}\)[/tex], we should follow these steps carefully:
1. Calculate the numerator:
- We have [tex]\((1.26 \frac{ g }{ mL }) \left(\frac{1 mL }{10^{-3} L }\right)\)[/tex].
- Note that [tex]\(1 mL\)[/tex] is equivalent to [tex]\(10^{-3} L\)[/tex]. Therefore, [tex]\(\frac{1 mL }{10^{-3} L } = 10^3\)[/tex].
- Multiplying [tex]\(1.26 \frac{ g }{ mL }\)[/tex] by [tex]\(10^3\)[/tex] gives:
[tex]\[
1.26 \times 10^3 = 1260 \; g/L
\][/tex]
2. Denominator:
- The denominator is [tex]\(44.49 \; \frac{ g }{ mol }\)[/tex], which remains as is.
3. Divide the numerator by the denominator:
- We want to find [tex]\(\frac{1260 \; g/L}{44.49 \; g/mol}\)[/tex].
- Dividing these two values:
[tex]\[
\frac{1260}{44.49} \approx 28.32 \; L/mol
\][/tex]
4. Assembly of the final result:
- The numerator is [tex]\(1260 \; g/L\)[/tex].
- The denominator is [tex]\(44.49 \; g/mol\)[/tex].
- The resulting quotient is approximately [tex]\(28.32 \; L/mol\)[/tex].
Thus, the detailed solution to the expression is:
- Numerator: [tex]\(1260.0\)[/tex]
- Denominator: [tex]\(44.49\)[/tex]
- Result: [tex]\(28.32097100472016\)[/tex]
So, the final answer is approximately [tex]\(28.32 \; L/mol\)[/tex].
