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Explain how you would prepare 50.0 mL of a 0.400 M solution o Potassium nitrate

ID: 539006 • Letter: E

Question

Explain how you would prepare 50.0 mL of a 0.400 M solution o Potassium nitrate using solid KNO_3. Do all necessary calculations and give enough detail that a student unfamiliar with the experiment could follow your directions. The MW of KNO_3 is 101.10 g/mol. If you use a beaker to make up the solution you will get a zero for this question. b. Explain how you would prepare 50.0 mL of a 0.200 M solution of potassium nitrate by diluting the 0.400 M solution of potassium nitrate. Do all necessary calculations and give enough detail that a student could follow your directions. c. Suppose that 25.0 mL of the 0.200 M solution of potassium nitrate is evaporated to dryness in an evaporating dish. What mass of potassium nitrate should be obtained?

Explanation / Answer

a) Molarity of solute = No of mole of solute per litre of solvent

   Mass of solute need for 1000ml = Molar mass × No of mole of solute

   Mass of solute for need for 50ml = (mass of solute need for 1000ml/1000ml)×50ml

Molarity of KNO3 = 0.400M

MW of KNO3 = 101.10g/mol

Mass of KNO3 for 1000ml = 0.400 mol × 101.10g/mol = 40.44g

Mass of KNO3 for 50ml = (40.44g/1000ml)×50ml = 2.022g

So, accurately weigh 2.022g of KNO3 and transfer it into a beaker and dissolve KNO3 using de ionised water. Transfer the solution present in beaker into a 50ml Volumetric flask and make up the solution to the mark using de ionised water

b) Apply the following formula

C1V1 = C1V1

Where,

C1 = Initial concentration ,0.400M

V1 = Initial Volume, ?

C2 = Final concentration ,0.200M

V2 = Final volume , 50ml

V1 = C2 × V2 / C1

= 0.200M × 50ml / 0.400M

= 25ml

So, Pipette out 25ml from 0.400M solution and transfer it into a 50ml Volumetric flask and make up it to the mark using de ionised water

C) Molarity of KNO3 solution = 0.200M

Volume of KNO3 solution = 25ml

No of mole of KNO3 = (0.200mol/1000ml)×25ml = 0.005mol

Mass of 0.005 mole KNO3 = 0.005mol × 101.10g/mol= 0.5055g

Therefore,

0.5055g of KNO3 should obtained

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