OP
Jaskanwar Singh
Aspiring Novelist
paul -
2nd -
Oxidation Number -
B
B
2nd -
8
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IIT-JEE problems
abhijangda
Padawan
I have seen some of the BMAT papers their level is slightly lower than that of FIITJEE AITS.
nims11
BIOS Terminator
@paul and jaskanwar
absolutely right, i cant believe i was trying to solve it using calculus
here's one more from yesterday's test
an equilibrium
P(Cl)5 <=> P(Cl)3 + (Cl)2
is obtained by starting with
1 mole of P(Cl)5 in a 1L vessel. vapour density of eqlb mixture was found to be 250.2
Noe, the eqlb mixture was transferred to a 2L vessel, its new degree of dissociation is x/100. what is the value of 'x' approx. (integer type)
one more,
two vertices of a triangle are (1,3) and (4,7). The orthocentre lies on the line x+y=3. find the locus of the third vertex.
absolutely right, i cant believe i was trying to solve it using calculus
here's one more from yesterday's test
an equilibrium
P(Cl)5 <=> P(Cl)3 + (Cl)2
is obtained by starting with
1 mole of P(Cl)5 in a 1L vessel. vapour density of eqlb mixture was found to be 250.2
Noe, the eqlb mixture was transferred to a 2L vessel, its new degree of dissociation is x/100. what is the value of 'x' approx. (integer type)
one more,
two vertices of a triangle are (1,3) and (4,7). The orthocentre lies on the line x+y=3. find the locus of the third vertex.
pauldmps
Banned
I'm trying your chemistry question but no success until now.
Only I was able to find a formula relating vapour density to degree of dissociation:
Degree of Dissociation = [(Dt - Do)/(n-1)Do] where 'Dt' is he theoretical VD (Molecular mass/2) & 'Do' is the observed VD (given in your question).
'n' is the number of molecules of product formed by dissociation of 1 molecule of reactant. In the case of your question, n = 2.
I was not able to find 'Dt' as I cannot find the eq. composition of the mixture. Try yourself.
Only I was able to find a formula relating vapour density to degree of dissociation:
Degree of Dissociation = [(Dt - Do)/(n-1)Do] where 'Dt' is he theoretical VD (Molecular mass/2) & 'Do' is the observed VD (given in your question).
'n' is the number of molecules of product formed by dissociation of 1 molecule of reactant. In the case of your question, n = 2.
I was not able to find 'Dt' as I cannot find the eq. composition of the mixture. Try yourself.
OP
Jaskanwar Singh
Aspiring Novelist
I tried maths. Nims -
x^2 + 31x + y^2 - 37y + 70 = 0
nims11
BIOS Terminator
Sorry for replying late guyz, i had my farewell today.
for maths,
the solution is
for the chemistry one, i am too stuck. I am not satisfied with solution given in FIITJEE solution.
isnt this relation supposed to be correct? -
Total moles initially * vapour density initially = Total moles equilibrium * vapour density equilibrium.
BTW, did anyone took a look at the exam special of this month's DIGIT DVD?
for maths,
the solution is
3x^2 + xy - 4y^2 - 2x + 24y - 40 = 0
i am not showing the whole method and just giving the basic idea.
take the point of a -> (x,y)
we can get the parametric form of the orthocentre (O) -> (a,3-a)
BO is perpendicular to AC, using the product of slopes, find a in terms of x and y.
using the product of slopes of AO and BC (which r perpendicular), an expression will be formed in terms of x,y,and a. put the value of a and the locus will be obtained.
i am not showing the whole method and just giving the basic idea.
take the point of a -> (x,y)
we can get the parametric form of the orthocentre (O) -> (a,3-a)
BO is perpendicular to AC, using the product of slopes, find a in terms of x and y.
using the product of slopes of AO and BC (which r perpendicular), an expression will be formed in terms of x,y,and a. put the value of a and the locus will be obtained.
for the chemistry one, i am too stuck. I am not satisfied with solution given in FIITJEE solution.
isnt this relation supposed to be correct? -
Total moles initially * vapour density initially = Total moles equilibrium * vapour density equilibrium.
BTW, did anyone took a look at the exam special of this month's DIGIT DVD?
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abhijangda
Padawan
math one is easy.
but regarding chem, i would say there is some error in values because I am getting degree of dissociation in first case as negative.
Also FIITJEE solutions error prone, some of questions contains error.
This solution is absolutely correct, just try to find Vapour density and total moles at equilibrium and see this equation is correct.
But still using this equation I am getting same negative degree of dissociation??
T
The Conqueror
Guest
Some Good Questions for Practice!
1.In solid state each ammonia molecule has 6 other ammonia molecules as nearest
neighbours. ΔH of sublimation of NH3 molecule at melting point is 30.2 kJ/mole
and in absence of H − bonding the ΔH sublimation is 14.4 kJ/mole. The strength
of H − bond in solid state is
(A) 166.8 kJ/mol (B) 55.6 kJ/mol
(C) 27.8 kJ/mol (D) 13.9 kJ/mol
2.Find out the number of atoms present in 100 g of the element if the element
occurs in bcc structure with edge length of 290 pm, density = 6.8 gcm−3.
(A) 6.023 × 1023 atoms (B) 12.046 × 1023 atoms
(C) 24.09 × 1023 atoms (D) 3.073 × 1023 atoms
3.If 2 moles of ideal gas at 27°C is subjected to expand reversibly ten times of its
initial volume, the change in entropy of expansion of the gas is
(A) 19.59 JK−1 mol−1 (B) 38.29 JK−1 mol−1
(C) 9.79 JK−1 mol−1 (D) 29.38 JK−1 mol−1
BTW, Khanacademy.org is just "WOW!". Superb Videos on Chemistry,Physics and Maths! Sal Khan is doing a wonderful job doing free videos and there are few IIT-JEE Videos too!. Unlike other sources where lectures are monotonic and scripted - this is not the case here!
1.In solid state each ammonia molecule has 6 other ammonia molecules as nearest
neighbours. ΔH of sublimation of NH3 molecule at melting point is 30.2 kJ/mole
and in absence of H − bonding the ΔH sublimation is 14.4 kJ/mole. The strength
of H − bond in solid state is
(A) 166.8 kJ/mol (B) 55.6 kJ/mol
(C) 27.8 kJ/mol (D) 13.9 kJ/mol
2.Find out the number of atoms present in 100 g of the element if the element
occurs in bcc structure with edge length of 290 pm, density = 6.8 gcm−3.
(A) 6.023 × 1023 atoms (B) 12.046 × 1023 atoms
(C) 24.09 × 1023 atoms (D) 3.073 × 1023 atoms
3.If 2 moles of ideal gas at 27°C is subjected to expand reversibly ten times of its
initial volume, the change in entropy of expansion of the gas is
(A) 19.59 JK−1 mol−1 (B) 38.29 JK−1 mol−1
(C) 9.79 JK−1 mol−1 (D) 29.38 JK−1 mol−1
BTW, Khanacademy.org is just "WOW!". Superb Videos on Chemistry,Physics and Maths! Sal Khan is doing a wonderful job doing free videos and there are few IIT-JEE Videos too!. Unlike other sources where lectures are monotonic and scripted - this is not the case here!
nims11
BIOS Terminator
3. B
1. The difference bet. the two 'H' will be the enthalpy of the h-bonds. One mole of ammonia wil hav 3 moles of H-bonds. But the answr is not matching.
2. I found the mass of a unit cell using the density and then divided it by 100. Then multiplying it by 2 shud giv the answer bt this too not matching.
Is there smthin wrong wid my methods?
1. The difference bet. the two 'H' will be the enthalpy of the h-bonds. One mole of ammonia wil hav 3 moles of H-bonds. But the answr is not matching.
2. I found the mass of a unit cell using the density and then divided it by 100. Then multiplying it by 2 shud giv the answer bt this too not matching.
Is there smthin wrong wid my methods?
OP
Jaskanwar Singh
Aspiring Novelist
Nims dont u think 1 ammonia molecule will have 4 hydrogen bonds. 3 hydrogen with other nitrogens and 1 nitrogen with 1 other hydrogen.
T
The Conqueror
Guest
The one video that I exceptionally liked was the Introduction to Atom. He has a unique way to teach without boring the students with fancy words and notations.
He has videos from basic addition to Advanced Calculus and Linear Algebra.
Pretty amazing and that too for free
What I really like is he wants us to understand the intuitive idea rather than just rote/memorize. . I guess no educational bureaucracy no matter how much you spend on it, will generate this outcome.
). I just hope that he uploads more videos related to our syllabus.
OP
Jaskanwar Singh
Aspiring Novelist
Thanks for link conquerer.
Nims u didnt ans my querry in previous post.
A doubt- (post solution too)
At the moment t=0 the force F=at is applied to a small body of mass m resting on a smooth horizontal plane. a is a constant. The permanent direction of this forms an angle 'd' with the horizontal. Find:
(a) the velocity of the body at the moment of its breaking off the plane;
(b) the distance traversed by the body up to this moment.
Nims u didnt ans my querry in previous post.
A doubt- (post solution too)
At the moment t=0 the force F=at is applied to a small body of mass m resting on a smooth horizontal plane. a is a constant. The permanent direction of this forms an angle 'd' with the horizontal. Find:
(a) the velocity of the body at the moment of its breaking off the plane;
(b) the distance traversed by the body up to this moment.
nims11
BIOS Terminator
for the physics doubt.
for your previous query, the answer is quite simple but i can find suitable words to explain it. as we are looking at many NH3 molecules, we need to remember that a hydrogen bond is shared between two atoms.all the hydrogen bonds will be between hydrogen and nitrogen so, to calculate the number of H-bonds in given number of moles, we just need to calculate number of hydrogen atoms or number of Nitrogen atoms multiplied by 3.
(horizontal)
a*t*cos(d)=m*(dv/dt)
integrating dv from 0 to 'v' and dt from '0' to 't', we get velocity as a function of time.
mv=a*cos(d)*(t^2) -----------------(1)
m(dx/dt)=a*cos(d)*(t^2)/2
again integrating ds from 0 to 'x' and dt from 0 to 't', we will get displacement(horizontal) as a function of time.
mx=a*cos(d)*(t^3)/6 ----------------------(2)
when the body breaks off the plane, t=mg/asin(d).
put this value of 't' in (1) and (2) to get the answer
a*t*cos(d)=m*(dv/dt)
integrating dv from 0 to 'v' and dt from '0' to 't', we get velocity as a function of time.
mv=a*cos(d)*(t^2) -----------------(1)
m(dx/dt)=a*cos(d)*(t^2)/2
again integrating ds from 0 to 'x' and dt from 0 to 't', we will get displacement(horizontal) as a function of time.
mx=a*cos(d)*(t^3)/6 ----------------------(2)
when the body breaks off the plane, t=mg/asin(d).
put this value of 't' in (1) and (2) to get the answer
for your previous query, the answer is quite simple but i can find suitable words to explain it. as we are looking at many NH3 molecules, we need to remember that a hydrogen bond is shared between two atoms.all the hydrogen bonds will be between hydrogen and nitrogen so, to calculate the number of H-bonds in given number of moles, we just need to calculate number of hydrogen atoms or number of Nitrogen atoms multiplied by 3.
OP
Jaskanwar Singh
Aspiring Novelist
OH MAN!!! i did a silly mistake in integration. didnt square the t!!!!!!!!!!!!!!!!!!!!!!!
but i said that wouldnt there be 4 HBs surrounding the single NH3 molecule. but i got your point of HBs in moles thanks. and thanks for physx correction too.
but i said that wouldnt there be 4 HBs surrounding the single NH3 molecule. but i got your point of HBs in moles thanks. and thanks for physx correction too.