The mass of Haemoglobin in mg required to pro | vedclass

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Question

(D) Gold number is defined as the minimum mass of a protective colloid in $mg$ that prevents the coagulation of $10 \ mL$ of a gold sol when $1 \ mL$

Vedclass · Accepted Answer

$0.15$

Vedclass · Answer

(D) Gold number is defined as the minimum mass of a protective colloid in $mg$ that prevents the coagulation of $10 \ mL$ of a gold sol when $1 \ mL$ of $10 \% NaCl$ solution is added to it. Given,the gold number of Haemoglobin $= 0.03$. This means $0.03 \ mg$ of Haemoglobin is required to protect $10 \ mL$ of gold sol against $1 \ mL$ of $10 \% NaCl$. For $50 \ mL$ of gold sol,the amount of protective colloid required is $5 	imes 0.03 \ mg = 0.15 \ mg$. Since the amount of $NaCl$ added is $5 \ mL$ (which is $5 	imes$ the standard amount),the protection required is proportional to the volume of the gold sol. Thus,the mass of Haemoglobin required is $0.15 \ mg$.

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 $= 0.03$)

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