- 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
14 - Solutions Questions Answers
Q.WRITE CELL RECTION OF THE FOLLOWING- .{1} ZN | ZN2+ || cu2+ |cu {2} PT {H2} | HCL || CL2{PT}
Zn -----------> Zn++ + 2e-
Cu++ + 2e- -----------> Cu
H2 --------> 2H+ + 2e-
Cl2 + 2e- --------> 2Cl-
one molar solution of sulphuric acid is equal to 2N. how???
sir as we know for sul.acid n factor = 2
so normality will be = 2xM thus M = N/2
but ans is 2N so plz give account of it .
molarity = 1
and N = nM = 2(1) = 2
hence it is clear
q. = Mass of solution of 1 molal glucose solution to get 0.2 molal glucose is.
a.200
b.300
c.236
d.108
I think question is incomplete, correct it first
acetic acid in a solvent undergoes 20% dissociation and 70% association, find vant hoff factor.
CH3COOH = CH3COO- + H+
1 0 0
0.8 0.2 0.2
CH3COO- + H+ = CH3COOH
0.2 0.2 0.8
0.06 0.06 0.8+0.14
0.06 0.06 0.94 sum = 1.06
so i = 1.06/1 = 1.06
calculate the molecular weight of cellulose acetate if its 0.2%wt/vol solution in acetone(sp. gravity 0.8) shows osmotic rise of 23.1mm against pure acetone at 27C.
By formula πV = nRT
π = hdg, n=w/M
given
0.2%wt/vol solution in acetone means w= 0.2 gm, V = 100 ml= 100/1000 litre = 0.1*10-3 m3 , h = 23.1/1000 m, d = 800 kg/m3 ,
now solve
gas in gas is a solution but interaction between the molecules play almost no role so these solutions are addressed as mixtures
WHAT OSMOTIC PRESSURE WOULD BE OF 1.25m SUCROSE SOLUTION AT 25C ?DENSITY OF SOLUTION IS 1.34g/mL?
formula is πV = nRT or π = MRT, M is molarity
strength of solution is 1.25m
so 1000 gm water ..............................................1.25 mole sucrose
so 1000 gm water .............................................1.25*342=427.5gm sucrose
so 1427.5 gm solution ....................................1.25 mole sucrose
so 1427.5/1.34=1065.30 ml = 1.0653 L solution ...........................1.25 mole sucrose
so M = 1.25/1.0653
put all the values to get answer.
THE PRESSURE OF ATMOSPHERE AIR IS 720 mm . THE PRESSURE OF WATER VAPOUR IN THE AIR IS
A) 5.55
B) 21
C) 18
D) ZERO
In this question atmospheric pressure should be given then
P0 - 720 mm will be the answer.
what will be the conc of nitrate ion when 400ml of 0.1 M AgNO3 is mixed with 100 ml of 0.2 m BaBr2
no. of miligm. eq. of AgNO3 = 400*0.1 = 40
no. of miligm. eq. of BaBr2 = 100*0.2*2 = 40
so no of miligm. eq. of Ba(NO3)2 = 40 so no. of milimoles of Ba(NO3)2 = 20
so [NO3-] = 20*2/500 = .08 M
WHICH OF THE FOLLOWING HAS LOWEST BOILING PT. (HARY. PMT 2003)
1) NaCl
2) CuCl
3) CuCI2
4) CsCl
SIR ANSWER GIVEN IS(3) . BUT AS THE NO. OF IONS IS MAX IN CuCl2 SO IT SHOULD HAVE MAX BP . THEN WHY ITS BP WILL BE MIN. ??
The only possibility is that the dissociation of the salt CuCl2 is low yet NaCl and CsCl are strongly dissociable salts. But CuCl should be less dissociable than CuCl2 . On this basis answer should be CuCl.
It is also probable that the question will be highest boiling point.