- 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
11 - 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??
“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
its 4th option
because As2S3 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 of the following have maximum value of enthalpy of physisorption?
(a) H2
(b) N2
(c) H2O
(d) CH4
(d) CH4 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 H2 , CH4 , CO2 , and NH3 absorbed by 1g of charcoal at 288K are in the order
H2 4.7 cc per gm
CH4 16.2 cc per gm
CO2 48 cc per gm
NH3 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
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
use t = 2.303/k * log(15.3/7)
k = 0.6932/t1/2
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
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
dn/dt = kn
so dn = (0.6932/24400*3.1*107)*(1/239)*6.023*1023*1
solve
a radioactive substance (M= 100g/mol) have t½ of 5 days. if today 0.125mg is left over what was its activity 20 days earlier?
(a) 522.8 Ci
(b) 1.93*10^7 Rd
(c) 1.93*10^13 Bq
(d) all of these
20 day means 4 half lifes so today 0.125mg means 0.125*24 = 2gm earlier
so activity dn/dt = kn
solve and convert in required units
a uranium mineral contains U-238 and Pb-206 in the ratio 4:1 by weight. Calculate the age of mineral (t½ for U-238 = 4.5 *10^9 years)
(a) 1.45*10^9 yrs
(b) 16.6*10^9 yrs
(c) 16.6*10^10 yrs
U:Pb = 4:1
taking these ratio as mass
we get the initial mass of U = 4+(1*235/208)
and present mass of U = 4
now use the first order equation of radioactivity