A laboratory assistant OK needs to prepare 225 mL of 0.150M CaCl2 solution how many grams of calcium chloride will she need?

First, we can determine how many moles of calcium chloride the lab assistant needs by multiplying the volume of the solution by its molarity.

M = n / V

where:

M = molarity
n = number of moles in solution
V = volume of solution (in Liters)

So:

[tex]V\cdot M = \not V \cdot\dfrac{n}{\not V} = n[/tex]

↓ plugging in the given values (with 225 mL converted to 0.225 L)

n = (0.225 L) · (0.150 M)

n = 0.03375 mole

Next, we can find the mass in grams of the calcium chloride needed by multiplying the moles needed (n) by CaCl₂'s molar mass:

[tex]m = (0.03375\text{ mole}) \left( \dfrac{40.08 + 2(35.45) \text{ g}}{1 \text{ mole}}\right)[/tex]

m = 3.74 g

So, the lab assistant will need 3.74 grams of calcium chloride to prepare a 225 mL solution that has a molarity of 0.150 M.