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The mass of haemoglobin in mg required t...

The mass of haemoglobin in mg required to protect from coagulation of 50 mL of a gold sol on adding 5 mL of 10 % NaCl solution is (gold number of haemoglobin = 0.03)

A

0.03

B

0.75

C

`0.30`

D

0.15

Text Solution

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The correct Answer is:
D
Gold number is the number of milligram of the propective colliod which prevents the coagulation of 10 mL of red gold solution. When 1 mL of a 10 percent solution of sodium chloride is added to it. Here, gold number of Hb = 0.03.
So. Mass of Hb require for 10 mL of gold is 0.03 mg. So, weight (mass) require for 50 mL of gold solution is 0.15 mg.
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