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.

Explanation:

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

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

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

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

$0.237=\frac{xg\times 1000}{229g/mol\times 500ml}$

$x=27.1$

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