In a group assignment, students are required to fill 10 beakers with [tex][tex]$0.720\, M\, CaCl_2$[/tex][/tex]. If the molar mass of [tex][tex]$CaCl_2$[/tex][/tex] is 110.98 [tex][tex]$g/mol$[/tex][/tex] and each beaker must have 250 [tex][tex]$mL$[/tex][/tex] of solution, what mass of [tex][tex]$CaCl_2$[/tex][/tex] would be used?

Use [tex]\text{molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}[/tex].

A. [tex]3.13\, g[/tex]
B. [tex]38.5\, g[/tex]
C. [tex]200\, g[/tex]
D. [tex]617\, g[/tex]



Answer :

Sure, let's work through the problem step-by-step:

1. Determine the volume of solution for one beaker in liters:

Each beaker needs 250 mL of solution. Since 1 liter (L) is equal to 1000 milliliters (mL), we can convert the volume from mL to L.

[tex]\[ \text{Volume per beaker} = \frac{250 \, \text{mL}}{1000 \, \text{mL/L}} = 0.25 \, \text{L} \][/tex]

2. Calculate the total volume of solution required for all beakers:

There are 10 beakers, each needing 0.25 L of solution.

[tex]\[ \text{Total volume} = 0.25 \, \text{L/beaker} \times 10 \, \text{beakers} = 2.5 \, \text{L} \][/tex]

3. Calculate the moles of CaCl_2 needed:

Using the molarity formula:

[tex]\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \][/tex]

We can solve for the moles of solute (CaCl_2):

[tex]\[ \text{moles of CaCl}_2 = \text{Molarity} \times \text{Total volume in liters} \][/tex]

Given the molarity (M) is 0.720 M,

[tex]\[ \text{moles of CaCl}_2 = 0.720 \, \text{M} \times 2.5 \, \text{L} = 1.8 \, \text{moles} \][/tex]

4. Calculate the mass of CaCl_2 needed:

Using the molar mass of CaCl_2, which is 110.98 g/mol, we can find the mass by multiplying the moles by the molar mass:

[tex]\[ \text{mass of CaCl}_2 = \text{moles of CaCl}_2 \times \text{molar mass of CaCl}_2 \][/tex]

[tex]\[ \text{mass of CaCl}_2 = 1.8 \, \text{moles} \times 110.98 \, \text{g/mol} = 199.76 \, \text{g} \][/tex]

Thus, the mass of CaCl_2 required is approximately 200 g.

So, the correct answer is:

[tex]\[ \boxed{200 \, \text{g}} \][/tex]