A student sets up the following equation to solve a problem in solution stoichiometry.
(The ? stands for a number the student is going to calculate.)

Enter the units of the student's answer.

[tex]\[
\frac{\left(1.10 \, \frac{\text{g}}{\text{mL}}\right)\left(\frac{1 \, \text{mL}}{10^{-3} \, \text{L}}\right)}{\left(65.99 \, \frac{\text{g}}{\text{mol}}\right)} = \text{?}
\][/tex]

[tex]\(\times 10\)[/tex]

[tex]\(\square\)[/tex]

[tex]\(\square - \square\)[/tex]



Answer :

Let's solve the given equation step-by-step to determine the units of the student's answer.

The equation given is:

[tex]\[ \frac{\left(1.10 \frac{ g }{ mL }\right)\left(\frac{1 mL}{10^{-3} L}\right)}{\left(65.99 \frac{ g }{ mol }\right)}=\text { ? } \][/tex]

Step 1: Identify the components of the equation.
- [tex]\(1.10 \frac{g}{mL}\)[/tex] is the density.
- [tex]\(\frac{1 mL}{10^{-3} L}\)[/tex] is the conversion factor from milliliters to liters.
- [tex]\(65.99 \frac{g}{mol}\)[/tex] is the molar mass.

Step 2: Analyze the units.
- The numerator [tex]\(\left(1.10 \frac{g}{mL}\right) \left(\frac{1 mL}{10^{-3} L}\right)\)[/tex] has units [tex]\(\frac{g}{mL} \times \frac{mL}{L}\)[/tex].

Since [tex]\(\frac{mL}{mL}\)[/tex] cancels out, leaving [tex]\(g/L\)[/tex], we are left with:

[tex]\[ \frac{1.10 \, g \times 10^{3} \, L^{-1}}{65.99 \frac{g}{mol}} \][/tex]

Step 3: Simplify the units obtained so far.
- [tex]\(g\)[/tex] cancels out since there is [tex]\(g\)[/tex] in both the numerator and denominator.

This leaves us with [tex]\( \frac{L^{-1}}{mol^{-1}} = \frac{mol}{L} \)[/tex].

So, the final units of the student's answer are [tex]\(mol/L\)[/tex].

Therefore, the student's answer would be:

[tex]\[ \boxed{mol/L} \][/tex]