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. 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

To calculate first let us define molarity

Molarity (M)= number of moles/ volume of solution (in litre)

                = weight of solute (in g)/( Mw of solute * Volume of solution (in litre)) ...........(1)

a) Molarity of solution given =0.4 M

   Volume of solution =50 ml =50 * 10-3 Litre

Mw of Potassium nitrate = 101.1 g/mol

subsituting all the above informaton in equation (1) to obtain the grams of potassium nitrate required .

0.4 =weight of solute/(101.1*50*10-3)

Weight of solute =0.4*101.1*50*10-3=2.022 grams

Hence we will dissolve 2.022 grams of solute in first say 20 ml of water to properly dissolve the solid solute and than further dilute the mixture to 50 ml

b) To calculate this we will use the following relation

      M1V1=M2V2    .............(2)

Given: M1 =0.2 M, V1=50 ml. M2=0.4 M

Substituting all the above values in equation (2)

0.2*50=0.4*V2

From the above equation we obtain V2=25 ml

Hence we will take 25 ml of 0.4 M potassium nitrate solution and than dilute the mixture to 50 ml to obtain 50 ml of 0.2 M potassium nitrate solution

c) Given : V=25 ml=25 *10-3 Litre

               M=0.2 m

Let us first calculate the amount of solut. i.e, potassium nitrate present in 25 ml of solution

        0.2=W/(101.1 *25*10-3)

W= 0.2*101.1*25*10-3 = 0.5055 grams

Hence, 0.5055 grams of potassium is present in the solution. Hence, when evaporation takes place and all of the water gets evaporated, 0.5055 grams of potassium nitrate should be obtained.

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