The coagulation of $100$ mL of a colloidal sol of gold is completely prevented by addition of $0.03$ g of haemoglobin to it before adding $1$ mL of $10 %$ $N a C l$ solution. Calculate the gold number of haemoglobin.

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $3$
Haemoglobin added to $100$ mL of gold sol to prevent coagulation by $1$ mL of $10 %$ $N a C l$ sol $= 0.03$ g $= 30$ mg.
Haemoglobin required to be added to $10$ mL of gold sol to prevent coagulation by $1$ mL of $10 %$ $N a C l$ sol $= 3$ mg.
Therefore, gold number of haemoglobin is $3$ .

Suggest Corrections

Similar questions

Q. The coagulation of 200 ml of a colloidal solution of gold is completely prevented by adding 0.50 g of starch to it before adding 1 ml of 10%

N a C l

solution. Calculate the gold number of starch

The coagulation of 100 ml of colloidal solution of gold is completely prevented by addition of 0.25 g of a substance “X” to it before adding 1 ml of 10% NaCl solution. The gold number of “X” is

Gold number of haemoglobin is $0.03$ . Hence, $100 m l$ of gold sol will require haemoglobin so that gold is not coagulated by $10 m L$ of $10 % N a C l$ solution:

Q. The coagulation of

200 m l

of a colloidal solution of gold is completely prevented by adding

0.50 g m

of starch it before adding

20 m l

10 % N a C l

solution. Calculate the gold number of starch.

Q. The coagulation of 100ml of colloidal solution of gold is completely prevented by addition of 0.25g of a substance of X to it before adding 1ml of 10% NaCl solution. the gold number of "X" is-

Join BYJU'S Learning Program

The coagulation of 100 mL of a colloidal sol of gold is completely prevented by addition of 0.03 g of haemoglobin to it before adding 1 mL of 10% NaCl solution. Calculate the gold number of haemoglobin.

The correct option is C 3Haemoglobin added to 100 mL of gold sol to prevent coagulation by 1 mL of 10% NaCl sol =0.03 g =30 mg.Haemoglobin required to be added to 10 mL of gold sol to prevent coagulation by 1 mL of 10% NaCl sol =3 mg.Therefore, gold number of haemoglobin is 3.

The coagulation of $100$ mL of a colloidal sol of gold is completely prevented by addition of $0.03$ g of haemoglobin to it before adding $1$ mL of $10 %$ $N a C l$ solution. Calculate the gold number of haemoglobin.