Question

calculate the no of millimoles contained in 500 mg of BacrO4​

Answer 1

Explanation:

Answer:

Millimoles\ of\ FeSO_4.C_2H_4(NH_3)_2SO_4.4H_2O= 1.3084\ millimolesMillimoles of FeSO4.C2H4(NH3)2SO4.4H2O=1.3084 millimoles

The mass of BaCrO_4BaCrO4 required = 5.0674 g

Explanation:

The formula for the calculation of moles is shown below:

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

Given :

Mass = 500 mg

Molar mass of FeSO_4.C_2H_4(NH_3)_2SO_4.4H_2OFeSO4.C2H4(NH3)2SO4.4H2O = 382.1455 g/mol

Thus,

Millimoles= \frac{500\ mg}{382.1455\ g/mol}Millimoles=382.1455 g/mol500 mg

Millimoles= 1.3084\ millimolesMillimoles=1.3084 millimoles

Also,

Molar mass of BaCrO_4BaCrO4 = 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}moles=Molar massMass taken

Thus,

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

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

Given that volume = 100 mL

Also,  

1\ mL=10^{-3}\ L1 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}Molarity=Volume of the solutionMoles of solute

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

x = 5.0674 g

The mass of BaCrO_4BaCrO4 required = 5.0674 g