Answer:
[tex]Millimoles\ of\ FeSO_4.C_2H_4(NH_3)_2SO_4.4H_2O= 1.3084\ millimoles[/tex]
The mass of [tex]BaCrO_4[/tex] required = 5.0674 g
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Given :
Mass = 500 mg
Molar mass of [tex]FeSO_4.C_2H_4(NH_3)_2SO_4.4H_2O[/tex] = 382.1455 g/mol
Thus,
[tex]Millimoles= \frac{500\ mg}{382.1455\ g/mol}[/tex]
[tex]Millimoles= 1.3084\ millimoles[/tex]
Also,
Molar mass of [tex]BaCrO_4[/tex]= 253.37 g/mol
Let the mass of the salt to prepare 100 mL 0.200 M solution = x g
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{x\ g}{253.37\ g/mol}[/tex]
[tex]Moles= \frac{x}{253.37}\ mol[/tex]
Given that volume = 100 mL
Also,
[tex]1\ mL=10^{-3}\ L[/tex]
So, Volume = 100 / 1000 L = 0.1 L
Molarity = 0.200 M
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]0.200=\frac{x}{253.37\times {0.1}}[/tex]
x = 5.0674 g
The mass of [tex]BaCrO_4[/tex] required = 5.0674 g
