Question

You need to make an aqueous solution of 0.237 M chromium(III) acetate for an experiment in lab, using a 500 mL volumetric flask. How much solid chromium(III) acetate should you add

Answer 1

Answer: 27.1 g of solid chromium(III) acetate should be added.

Step-by-step explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

`Molarity=(n* 1000)/(V_s)`

where,

n = moles of solute

`V_s` = volume of solution in ml

moles of
`Cr(CH_3COO)_3` =
`\frac{\text {given mass}}{\text {Molar mass}}=(xg)/(229g/mol)`

Now put all the given values in the formula of molality, we get

`0.237=(xg* 1000)/(229g/mol* 500ml)`

`x=27.1`

Therefore, 27.1 g of solid chromium(III) acetate should be added.