Question

Calculate the number of millimoles contained in 500mg of FeSO4•C2H4(NH3)2SO4.4H20 Calculate the number of grams of BaCrO4 that would have to be dissolved and diluted to 100ml to prepare a 0.200M solution.

Answer 1

Answer:

$Millimoles\ of\ FeSO_4.C_2H_4(NH_3)_2SO_4.4H_2O= 1.3084\ millimoles$

The mass of $BaCrO_4$ required = 5.0674 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given :

Mass = 500 mg

Molar mass of $FeSO_4.C_2H_4(NH_3)_2SO_4.4H_2O$ = 382.1455 g/mol

Thus,

$Millimoles= \frac{500\ mg}{382.1455\ g/mol}$

$Millimoles= 1.3084\ millimoles$

Also,

Molar mass of $BaCrO_4$= 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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{x\ g}{253.37\ g/mol}$

$Moles= \frac{x}{253.37}\ mol$

Given that volume = 100 mL

Also,  

$1\ mL=10^{-3}\ L$

So, Volume = 100 / 1000 L = 0.1 L

Molarity = 0.200 M

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$0.200=\frac{x}{253.37\times {0.1}}$

x = 5.0674 g

The mass of $BaCrO_4$ required = 5.0674 g