300 mL of gold sol is mixed with 30 mL of 10% NaCl solution. The mass of haemoglobin in mg required to protect the gold sol from coagulation is (gold number of haemoglobin is 0.03)

Options

A. 0.3
B. 0.09
C. 0.03
D. 0.9

Answer

D

Explanation

Gold number is the number of milligram of the protective colloid, which prevents the coagulation of 10 m.l of a gold sol on adding 1 mL of 10% NaCl solution. 30 mL of 10% NaCl requires 300 mL, coagulates 0.-3 mg, every 10 mL of 10% NaCl gives 0.03 mg. So, mass of haemoglobin required = 0.03(30) = 0.9 mg