Please go through how to solve each of these problems with the steps you used an

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Please go through how to solve each of these problems with the steps you used and how you derived your final answer. Thank you!

4. Describe the preparation of 500 mL of a 0.1 N H2SO4 assuming that you have in stock 18.0 M H2SO4. You may assume that this solution will be used for an acid-base volumetric analysis.

5. A spectroscopic method for determining albumin content in serum involves the preparation of a stock solution of albumin (6g/100mL aqueous solution). From the prepared stock, you need to prepare a series of working standard solutions. Assume reagent cost is a consideration so you plan to prepare a 1mL of each of these standards: 12, 24, 36, and 48 g/L. To complete this task you have 500 ?L and  1000?L automatic transfer pipets. Describe how you might prepare these standards.

Describe the preparation of 500 mL of a 0.1 N H2SO4 assuming that you have in stock 18.0 M H2SO4. You may assume that this solution will be used for an acid-base volumetric analysis. A spectroscopic method for determining albumin content in serum involves the preparation of a stock solution of albumin (6g/100mL aqueous solution). From the prepared stock, you need to prepare a series of working standard solutions. Assume reagent cost is a consideration so you plan to prepare a 1mL of each of these standards: 12, 24, 36, and 48 g/L. To complete this task you have 500 ?L and 1000?L automatic transfer pipets. Describe how you might prepare these standards.

Explanation / Answer

4)

0.1 N H2SO4 = 0.1/2 M H2SO4 = 0.05 M H2SO4 = 0.05 moles / L ....(as H2SO4 has 2 H+)

that is 0.05/2=0.025 moles in 500 ml

18 M = 18 moles per Litre

we need 0.025 moles so,

volume needed is x,

18/1=0.025/x

x = 0.025/18=0.001389 L = 1.389 ml

so,

Volume of water = 500-1.389 = 498.611 ml

so,

preparation : 498.611 ml of water + 1.389 ml of stock solution

5)

stock soltion : 6g/100 mL = 60 g/L

a)

12 g/L = 12 gram / Litre

so,

lets x gram per ml

so,

x = 12/1*10^-3 = 0.012 grams

so,

stock solution needed = 0.012/60 = 0.0002 L = 0.2 ml

so,

Preparation : 0.2 ml of stock solution + 0.8 ml of water

b)

24 g/L = 24 gram / Litre

so,

lets x gram per ml

so,

x = 24/1*10^-3 = 0.024 grams

so,

stock solution needed = 0.024/60 = 0.0004 L = 0.4 ml

so,

Preparation : 0.4 ml of stock solution + 0.6ml of water

48 g/L = 48 gram / Litre

so,

lets x gram per ml

so,

x = 48/1*10^-3 = 0.048 grams

so,

stock solution needed = 0.048/60 = 0.0008 L = 0.8 ml

so,

Preparation : 0.8 ml of stock solution + 0.2 ml of water

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