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Physical Chemistry
Q1. Calculate molarity of water, if its density is 1000 kg/m3 . (IIT 2003) Ans. 55.6
Q2. 1g of charcoal adsorbs 100 ml of 0.5 M CH3COOH to form monolayer and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 X 102 m2/kg. (IIT 2003) Ans. 114.72
Q3. A mixture of 20ml of CO,CH4 and N2 was burnt in excess of O2 resulting in reduction of 13ml of volume. The residual gas was treated with KOH solution to show a contraction of 14ml in volume. Calculate volume of CO,CH4 and N2 in mixture. All measurements are made at constant pressure and temperature. (IIT 1995) Ans. 10,4,6
Q4. When a mixture of 10 mole of SO2, 15 mole of O2 was passed over catalyst , 8 mole of SO3 was formed. How many mole of SO2 and O2 did not enter into combination? Ans. 2,11
Q5. 5ml of a gaseous hydrocarbon was exposed to 30 ml of O2. The resultant gas, on cooling is found to measure 25ml of which 10 ml are absorbed by NaOH and the remainder by pyrogallol. Determine molecular formula of hydrocarbon. All measurements are made at constant pressure and temperature. Ans C2H4
Q6. One litre of a mixture of CO and CO2 is passed through red hot charcoal in tube. The new volume becomes 1.4 litres. find out % composition of mixture by volume. Ans.40%,60%
Q7. Copper forms two oxides. For the same amount of copper, twice as much oxygen was used to form first oxide than to form second one. What is the ratio of the valencies of copper in first and second oxides? Ans.2:1
Q8. Haemoglobin contains 0.25% iron by weight. The molecular weight of haemoglobin is 89600. Calculate the no. of iron atoms per molecule of haemoglobin. Ans.4
pls answer with full and detailed explanations......
Comments (9)
ans 3)take volume of CH4 = x
N2 = y
CO = z
we are burning in excess of O2 so write the reactions firstly
CH4 + 2O2 -------> CO2 + 2H2O
CO + 1/2 O2 -------> CO2
N2 + O2 wont give u any reaction
we are given x + y + z = 20 ml.............a)
from x ml of CH4 we will get x ml of CO2
N2 will be unreacted
from z ml of CO we will get z ml of CO2 ,
KOH will react with CO2 (as KOH abosorbs carbon di oxide), so if it will abosorb , the volume will become less
it is given derz a contraction of 14 ml
so
x+z = 14 ml...............b)
now see the reactions
x ml of methane will take 2x ml of O2 (2 moles)
z ml of CO will take z/2 ml of O2 (1/2 moles)
total moles of O2 are given 13ml
so
2x + (z/2) = 13................c)
solve these equations......
No of moles of acetic acid initially present=MV/1000=0.5X100/1000=0.05
No of moles of acetic acid left=MV/1000=0.49X100/1000=0.049
No of moles of acetic acid adsorbed=0.05-0.049=0.001 mol.
No of molecules of acetic acid adsorbed=0.001x6.023x10^23=6.023x10^20
Area occupied by a single molecule of acetic acid=Total area/No of molecules adsorbed=(3.01x10^2)/6.023x10^20=5x10^ -19 M2
Let there be X Lt of CO in the mixture,hence,there will be (1-x)Lt of CO2.
CO2 only reacts with red hot charcoal.
This is given by the eqn
CO2(g) + C(S)---------------------> 2 CO(g)
1 vol ---------------------------------------- 2 vol
(1-x)------------------------------------------2(1-x)
In the reaction chamber only CO is present.
Total volume of gas becomes=X +2(1-X)=1.4 Lts
Hence X=0.6 Lt,So the % composition of CO=60%
The % composition of CO2 is 40%
jr.mohanty sir,
in Q6, how volume of 2CO becomes 2 tht i want to knw
In Q2, moles of acetic acid left is calculated as 0.49X100/1000 . . . how u have taken volume 100ml as 100ml was volume adsorbed by charcoal so volume left should be different...
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1m3 =1000 L , 1000L CONTAINS 1000KG OF WATER = 1000L contains 55.55 x 103 moles of water
therefore 1 mole of water contains 55.555 moles ,thus molarity = 55.555