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

Question

Explain how you would prepare 50.0 mL of a 0.400 M solution of 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. 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

moles of potasium nitrate (KNO3)=molarity* Volume in L= 0.4*50/1000 =0.02

molar mass of KNO3= 101 g/mole, mass of KNO3= moles* molar mass= 0.02*101 =2.02 gm

when 2.02 gm of KNO3 is dissolved in 50ml then the molarity becomes moles/liter= 2.02/(101*50/1000)= 0.4M

2. moles in 50ml of 0.2M= Molarity* volume(L)=0.2*50/1000 =0.01

let the volume of 0.4M be V1 (L)

hence moles in 0.4M=0.4*V1=0.01

V1=0.01/0.4= 0.025L= 0.025*1000ml=25 ml

25 ml of 0.4M solution 50ml gives 0.2M

3. moles of KNO3 in 25ml of 0.2M= 0.2*25/1000 =0.005moles

When this is evaporated all the water evaporates and 0.005 moles of KNO3 remains. Mass of KNO3= moles* molar mass =0.005*101=0.505 gm

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