2. Describe how you would prepare 100ml of 0.08 M sulfuric acid [H2SO4] from an

Question

2. Describe how you would prepare 100ml of 0.08 M sulfuric acid [H2SO4] from an 18 M concentrated sulfuric acid stock solution.

3. Describe how you would prepare 50 ml of 0.2 M potassium iodate [KIO3] dissolved in the 0.08 M sulfuric acid [H2SO4] that you have prepared above.

4. Describe how you would prepare 25 ml of 3.6 M hydrogen peroxide [h2O2] from a 30% (w/w) stock H2O2 solution (

5. Describe how you would prepare 50 ml of a 3% (w/v) soluble starch solution in water.

Molecular weights:

Malonic acid: 104.6

Manganese sulfate: 169.01

Potassium iodate: 214

Hydrogen peroxide: 34.01

Explanation / Answer

2. M1V1 = M2V2

18 X V1 = 0.08 X 100

V1 = 0.44mL

we will take 0.44mL of stock sulphuric acid and will make it up to 100mL to preapre the above solution

3.

mol wt of KIO3 = 214

so for 0.2M = mass / mol wt X volum in litres = mass / 214 / 0.05

so mass = 0.2 X 214 X 0.05 = 2.14g of =KIO3 will be dissolved in above sulphuric acid solution.

for 50mL of 0.2M we will

4. Find out the molarity for 30wt% Hydrogen peroxide,

Molarity= (30/100) * density * 1000 /( Molar mass hydrogen peroxide)
= 0.3 * 1.463 * 1000 / 34.0147
= 12.90 M
30 wt% hydrogen peroxide = 12.90 M hydrogen peroxide

Then, using M1V1=M2V2,
M1= 12.90 M (concentration in bottle)
V1= ? (as you don't know how much volume needed to take)
M2= 3.6M ( concentration required)
V2= 50 mL (volume required)

so,
12.90 V1= 3.6 *50
V1= 3.6 * 50 /12.90
= 13.95 mL

13.95 mL needed to take to prepare the solution

5. for 50mL of starch solution we will take 48.5mL of water and will add 1.5g of starch in it

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