The coagulation of 100 mL of gold sol is prevented by addition of 0.25 g of starch to it before adding 1 ml of 10% NaCl solution. If the gold number of starch is x then $\frac{x}{5}$ is: ___

Open in App

Solution

100 ml of gold sol needs = 250 mg of starch

10 ml of gold sol needs $= \frac{250}{10} = 25$ mg

So, x =25

$\frac{x}{5} = 5$

Suggest Corrections

Similar questions

Q. The coagulation of

10 {cm}^{3}

of Gold sol by

1 mL

10 %

N a C l

solution is completely prevented by addition of

0.25 g

of starch to it. The gold number of starch is:

Q. Gold number of potato starch is

25

. Weight of potato starch required to prevent coagulation of

100 m l

of gold sol by the addition of

1 m l

10 % N a C l

Q. The coagulation of

100 m L

of a colloidal solution of gold is completely prevented by adding

0.25 g

of starch to it before adding

10 m L

10

N a C l

solution. Find out 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-

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

Join BYJU'S Learning Program

The coagulation of 100 mL of gold sol is prevented by addition of 0.25 g of starch to it before adding 1 ml of 10% NaCl solution. If the gold number of starch is x then x5 is: ___

100 ml of gold sol needs = 250 mg of starch 10 ml of gold sol needs =25010=25 mg So, x =25 x5=5

The coagulation of 100 mL of gold sol is prevented by addition of 0.25 g of starch to it before adding 1 ml of 10% NaCl solution. If the gold number of starch is x then $\frac{x}{5}$ is: ___

100 ml of gold sol needs = 250 mg of starch

10 ml of gold sol needs $= \frac{250}{10} = 25$ mg

So, x =25

$\frac{x}{5} = 5$