Question

a student sets up and solves the following equation to solve a problem in solution stoichiometry. fill in the missing part of the students equation.
\\(\frac{(0.72)(\frac{1\text{ ml}}{10^{- 3}\text{ l}})}{(\frac{29.14\text{ g}}{\text{mol}})} = 25.\frac{\text{mol}}{\text{l}}\\)

Explanation:

Step1: Analyze the units and stoichiometry concept

We know that in solution - stoichiometry, we use conversion factors to relate different quantities. The right - hand side has units of $\frac{mol}{L}$ (molarity). The left - hand side has a mass - to - mole conversion factor $\frac{29.14\ g}{mol}$ and a volume conversion factor $\frac{1\ mL}{10^{-3}\ L}$. We need to find the missing part such that the units work out correctly.
Let the missing part be $x$. The overall equation should have the units balance.
The units of the left - hand side:
The $\frac{1\ mL}{10^{-3}\ L}$ converts volume from mL to L. The $\frac{29.14\ g}{mol}$ is a molar - mass conversion factor.
We know that molarity $M=\frac{n}{V}$, where $n$ is the number of moles and $V$ is the volume in liters.
We start with $\frac{(0.72)(\frac{1\ mL}{10^{-3}\ L})x}{(\frac{29.14\ g}{mol})}=25\frac{mol}{L}$
We want to isolate $x$. First, cross - multiply:
$(0.72)(\frac{1\ mL}{10^{-3}\ L})x=(25\frac{mol}{L})\times(\frac{29.14\ g}{mol})$
The $\frac{1\ mL}{10^{-3}\ L}$ is just a unit conversion factor for volume (1 mL = $10^{-3}$ L).
We know that $M = \frac{n}{V}$, and we can rewrite the equation in terms of mass and molar mass.
Let's work on the units:
The right - hand side has units of $\frac{mol}{L}\times\frac{g}{mol}=\frac{g}{L}$
The left - hand side should also have units of $\frac{g}{L}$.
If we assume the missing part is a mass in grams, let's solve for it.
Cross - multiplying gives us:
$x=\frac{(25\frac{mol}{L})\times(\frac{29.14\ g}{mol})\times(10^{-3}\ L)}{0.72}$

Step2: Calculate the value of $x$

First, simplify the numerator: $(25\frac{mol}{L})\times(\frac{29.14\ g}{mol})\times(10^{-3}\ L)=25\times29.14\times10^{-3}\ g$
$25\times29.14\times10^{-3}=25\times0.02914 = 0.7285\ g$
Then, $x=\frac{0.7285\ g}{0.72}\approx1.01\ g$

Answer:

$1.01$ g