The coagulation of $10 {cm}^{3}$ of gold sol by $1 mL$ 10% $N a C l$ solution is completely prevented by addition of $0.025 g$ of starch to it. The gold number of starch is:

Open in App

Solution

$0.025 g$ starch add in $10 {cm}^{3}$ gold sol
$∴ 10 {cm}^{3} gold sol contain = 0.025 g$
$1 {cm}^{3} gold sol contain = \frac{0.025 \times 10^{- 3} mg \times 10}{100} = 25$

Suggest Corrections

Similar questions

On addition of 0.5 mL of solution of 10% NaCl to 20 mL of red gold sol in presence of 0.025 g of starch, the coagulation is just prevented. The gold number of starch is

Q. On addition of one mL of

10 %

N a C l

solution to

10

mL gold sol in the presence of

0.25

g of starch, the coagulation is just prevented. Starch has gold number:

Q. On addition of one

1 m L

of solution of

10 %

N a C l

10 m L

of red gold sol in presence of

0.025 g

of starch, the coagulation is just prevented. The gold number of starch.

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: ___

Q. On addition of one

m L

10

N a C l

solution to

10 m L

gold sol in presence of

0.25 g m

of starch, the coagulation is just prevented, starch has gold number:

Join BYJU'S Learning Program

The coagulation of 10 cm3 of gold sol by 1 mL 10% NaCl solution is completely prevented by addition of 0.025 g of starch to it. The gold number of starch is:

0.025 g starch add in 10 cm3 gold sol ∴10 cm3gold sol contain=0.025 g 1 cm3gold sol contain=0.025×10−3 mg×10100=25

The coagulation of $10 {cm}^{3}$ of gold sol by $1 mL$ 10% $N a C l$ solution is completely prevented by addition of $0.025 g$ of starch to it. The gold number of starch is:

$0.025 g$ starch add in $10 {cm}^{3}$ gold sol
$∴ 10 {cm}^{3} gold sol contain = 0.025 g$
$1 {cm}^{3} gold sol contain = \frac{0.025 \times 10^{- 3} mg \times 10}{100} = 25$