Question

A student has 540.0 mL of a 0.1035 M aqueous solution of Na2CrO4 to use in an experiment. She accidentally leaves the container uncovered and comes back the next week to find only a solid residue. The mass of the residue is 19.12 g. Determine the chemical formula of this residue.

Answer 1

Answer:

The chemical formula of this residue: $Na_2CrO_4.10H_2O$

Explanation:

Mass of residue = 19.12

Let the formula of the residue be $Na_2CrO_4.xH_2O$

Moles of residue: n

$n=\frac{19.12}{162 g/mol+x\times 18 g/mol}$

Moles of sodium chromate = n'

Molarity of sodium chromate = 0.1035 M

Volume of sodium chromate solution = 540.0 mL = 0.540 L

1 mL = 0.001 L

$Concentration=\frac{moles}{Volume(L)}$

$0.1035 M=\frac{n}{0.540 L}$

$n'=0.1035 M\times 0.540 L=0.05589 mol$

$Na_2CrO_4+xH_2O\rightarrow Na_2CrO_4.xH_2O$

According to reaction,1 mole of  $Na_2CrO_4$ gives 1 mole of$Na_2CrO_4.xH_2O$

So, n = n'

$\frac{19.12}{162 g/mol+x\times 18 g/mol}=0.05589 mol$

x = 10

The chemical formula of this residue: $Na_2CrO_4.10H_2O$