- Organic Chemistry
- Aldehydes and Ketones
- Alkyl Halides, Alcohols and ethers
- Amines and other nitrogen compounds
- Aromatic Chemistry
- Carbohydrates, Amino acids, protein, Vitamin and Fat
- Carboxylic acids and its derivatives
- Chemistry in daily life
- General Mechanism in organic compounds
- Hydrocarbons
- Nomenclature and isomerism

# 16 - Surface Chemistry Questions Answers

0.25 g lyophilic colloid is added to 100 ml gold solution to prevent the coagulation on adding 1ml 10% NaCl soln . What will be gold number of lyophilic colloid??

**Asked By: SARIKA SHARMA**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

“weight of the dried protective agent in milligrams, which when added to 10 ml of a standard gold sol (0.0053 to 0.0058%) is just sufficient to prevent a colour change from red to blue on the addition of 1 ml of 10 % sodium chloride solution, is equal to the gold number of that protective colloid.”

0.25 g lyophilic colloid is added to 100 ml gold solution

so 250 mg lyophilic colloid is added to 100 ml gold solution

so 25 mg lyophilic colloid is added to 10 ml gold solution

so answer will be 25

Most effective coagulant for a colloid soln of arsenic sulphide in water is

1) 0.1 M sodium phosphate

2) 0.1M zinc sulphate

3) 0.1 M zinc nitrate

4) 0.1 M aluminium chloride

Plzz annex explanation also

**Asked By: SARIKA**19 Day ago

**Asked By: SARIKA SHARMA**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

its 4th option

because As_{2}S_{3 }is usually a negative colloid and positive ion is responsible for its coagulation and according to Hardy Schulze priciple more positive ion will be more effective

which gas is absorbed easily at solid surfaxe?

(a) SO_{2}

(b) H_{2}

(c) O_{2}

(d) N_{2}

**Asked By: HIMANSHU MITTAL**7 year ago

**Solved By: SARIKA**

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H_{2} gas is absorbed on a metal surface like platinum. this follow

(a) first order rx.

(b) second order rx.

(c) third order rx.

(d) 0 order rx.

**Asked By: HIMANSHU MITTAL**7 year ago

**Solved By: SARIKA**

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which of the following have maximum value of enthalpy of physisorption?

(a) H_{2}

(b) N_{2}

(c) H_{2}O

(d) CH_{4}

**Asked By: HIMANSHU MITTAL**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

(d) CH_{4} based on the value of energy which is 6 Kcal/mole for 0.1 milimole/gram and is greatest among the given 4

volume of gases H_{2} , CH_{4} , CO_{2} , and NH_{3} absorbed by 1g of charcoal at 288K are in the order

**Asked By: HIMANSHU MITTAL**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

H_{2} 4.7 cc per gm

CH4 16.2 cc per gm

CO_{2 } 48 cc per gm

NH_{3} 181 cc per gm

there are two radio nuclei A and B. A is α-emitter and B is a β-emitter, their decay constant are in the ratio 1:2. what should be the no. of atoms of A&B at time t=0, so that probability of getting α and β-particles are same at time t=0

(a) 2:1

(b) 1;2

(c) 1:4

(d) 4:1

**Asked By: HIMANSHU MITTAL**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

its (b)

decay constant ratio = 1:2

so half life ratio = 2:1

so for getting probability of finding alpha and beta same ratio will be 1:2

a carbon sample from the frame of a picture gives 7 counts of C-14 per minute per gram of carbon . if freshly cut wood gives 15.3 counts of C-14 per minute. calculate the age of frame (t½ of C-14 = 5570 yrs)

(a) 6286 yrs

(b) 5527 yrs

(c) 5570 yrs

(d) 4570 yrs

**Asked By: HIMANSHU MITTAL**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

use t = 2.303/k * log(15.3/7)

k = 0.6932/t_{1/2}

the binding energy per nucleon for C-12 is 7.68 MeV and that for C-13 is 7.47 MeV. calculate the energy required to remove a neutron from C-13

(a) 4.5MeV

(b) 45MeV

(c) 0.45MeV

(d) 0.045MeV

**Asked By: HIMANSHU MITTAL**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

7.47*13-7.68*12 = 97.11-92.16 = 4.5 Mev approx

1gm of 94Pu-239 is an α-emitter with a half life of 24400 yrs. given 1g sample of this plutonium how many α-particles will it emit per second?

(a) 2.27*10^9

(b) 2.27*10^10

(c) 2.27*10^11

(d) 2.27*10^12

**Asked By: HIMANSHU MITTAL**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

dn/dt = kn

so dn = (0.6932/24400*3.1*10^{7})*(1/239)*6.023*10^{23}*1

solve