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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]\[
\frac{(0.63 \, \text{mol})\left(\frac{1 \, \text{mL}}{10^{-3} \, \text{L}}\right)}{\left(4.0 \, \frac{\text{mol}}{\text{L}}\right)} = ?
\][/tex]

[tex]$\square$[/tex]


Sagot :

To determine the units of the student's answer, let's analyze the given equation step-by-step:

[tex]\[ \frac{(0.63 \text{ mol}) \left(\frac{1 \text{ mL}}{10^{-3} \text{ L}}\right)}{\left(4.0 \frac{\text{ mol}}{\text{ L}}\right)} = ? \][/tex]

First, let's clarify the units involved:

1. [tex]\(0.63 \text{ mol}\)[/tex] represents the amount of substance in moles.
2. [tex]\(\left(\frac{1 \text{ mL}}{10^{-3} \text{ L}}\right)\)[/tex] is a conversion factor; converting milliliters (mL) to liters (L). Since [tex]\(1 \text{ mL} = 10^{-3} \text{ L}\)[/tex], this factor effectively multiplies the number by [tex]\(10^3\)[/tex] and is dimensionless in this context.
3. [tex]\(4.0 \frac{\text{ mol}}{\text{ L}}\)[/tex] is the concentration in moles per liter (mol/L).

Now, plug these values into the equation and focus on the units:

[tex]\[ \frac{(0.63 \text{ mol}) \times (10^3)}{4.0 \frac{\text{ mol}}{\text{ L}}} \][/tex]

Next, handle the units step-by-step:

1. Multiply [tex]\(0.63 \text{ mol}\)[/tex] by [tex]\(10^3\)[/tex], giving:

[tex]\[0.63 \text{ mol} \times 10^3 = 630 \text{ mol} \cdot \text{ mL} / \text{ L}\][/tex]

2. Then divide by the concentration [tex]\(4.0 \frac{\text{ mol}}{\text{ L}}\)[/tex]:

[tex]\[\frac{630 \text{ mol}}{4.0 \frac{\text{ mol}}{\text{ L}}} = 157.5 \text{ L}\][/tex]

Notice that the moles ([tex]\(\text{ mol}\)[/tex]) in the numerator and denominator cancel out, leaving us with the unit of liters (L):

[tex]\[ \frac{\text{ mol}}{\text{ mol} / \text{ L}} = \text{ L} \][/tex]

So, the final result is in units of liters (L). Therefore, the student's answer should have the units:

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