1equivalent molecular wight of mixture =10
required ration is√(32/10)
hey lets start a thread for the prep of bitsat....will be posting in questions that might be of some relevence to the exam...
guys...plz post in ur questions
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127 Answers
1the first one goes like this....
A and B of equal volumes hold 3 mol of nitrogen + 2 mol of He and 5 mol of oxygen respectively at the same temperature.
both comtainers are kept separately in vacuum. Calculate the ratio of rates of diffusion of gas mixture in container A to that of oxygen if the area of orifice in both containers is same.
1i think what ur mistake is...may be u calculating the wt of the mixture inside the container.....but what wae need is the wt, of the mixture effusing out of the container
1uum..........then data is insuffiecient ............. OR rather if u knw....temme how to go abt the sum...........or wht is the answer????
1easy one!
we just find MOLECULAR WEIGHT OF MIXTURE=12.69
then simply use GRAHAM"S LAW which gives required ratio as √32/12.69=1.588
1ya heaven on earth is right....
vaise....can u people post more questions
1one mole of a mixture of CO and CO2 requires exactly 20 grams of NaOH to convert all the CO2 into NaCO3. How many more grams of NaOH would it require for conversions into Na2CO3 if the mixture ( 1 mole) is completely oxidized to CO2?
1Q1)Carbon fibre is made up of which polymer?
Q2) structure of malachite green??
Do they ask such stuff??
1Cute cat since i have a problem in tht chap ...posting one
1.> 1.2 g of commercial sample of oxalic acid was dissolvd in 200 ml. 10ml of sample reqiured 8.5 ml of N/10 KMnO4.Calc % puriity of sample [not getting the answer]
2>IONIC:
A solution has 0.09M HCl ,0.09M CHCl2COOH and 0.1M acetic acid .pH of soln is 1. If pka for acetic acid is 10-5 .then calc the pka of CHCl2COOH
11. If alpha and beta are the roots of the equation ax^2 + bx + c=0, the k lies between alpha and beta, if
a. ak^2 + bx + c < 0
b. a^2k^2 + abk + ac < 0
c. a^2k^2 + abk + ac>0
2. If the roots of equation (x-b)(x-c) + (x-c)(x-a) + (x-a)(x-b) = a are equal, then
a. a + b + c=0
b. a + bw + cw^2 = 0
c. a – b + c = 0
3. If x^2 + px + qr = 0, x^2 + rx + pq =0 has a common root, then product to there common roots is
a. pqr
b. 2pqr
c. (pqr)^2
4. The number of ways in which mn students can be distributed equally among m sections is?
5. The dew point and the atmospheric temperature on a particular day is found to be 18°C and 25C. If the saturation vapour pressure at 18 C and 25C be 15.5 mm and 25.7 mm of Hg, then its relative humidity is
a. 90%
b. 30.0%
c 60.3 %
P.S. - i know all the answers, i need the solutions
1glass is solubel in
a HClO4
b HF
c aqua regia
d H2SO4
ans is b....i guessed right...bt y is it so?
23@ cutecat
Q1> b
for f(x) = 0
coeff of x2 is a
we cant say whether a > 0 or < 0
hence we multiply f(x) by a
if alpha beta are roots of f(x) = 0 , they are roots of af(x)= 0 also
now coeff of x2 of the quadtratic af(x) is positive i.e a2, hence it is a parabola wich opens upwards,
hence for any point k between alpha beta, af(k) < 0
Q5> c ( do u need an explanation )
111) 
so a>0 and f(k) < 0
==> af(k) <0
==>a(ak2 + bk c )<0
==>a2k2 + abk + ac < 0
====> ans is b
---------------------------------------------------------------------------------
now if a < 0
f(k) > 0
===> af(k) < 0
which yeilds the same answer only...
11For that glass question...
SiO2 + 4HF -------------> SiF4 + 2H2O
11Q3) a
Let roots of x2 + px + qr = 0 be \alpha & \beta
Let roots of x2 + rx + pq = 0 be \alpha & \gamma
so \alpha\beta = q r .....(1)
& \alpha\gamma = p q.......(2)
now subtracting the two eqns....
x2 + px + qr = 0
- x2 + rx + pq = 0
-------------------------------------
x(p-r) + q(r-p) = 0
or (p-r)(x-q)=0
or x = q ......(common root...)
so \alpha = q
now comparing eqn 1 and 2
we get....
\beta = r
and \gamma = p
so \alpha\beta\gamma = pqr...
1q 1 and 5, i understood.....and that glass one also...thanks a lot
@ qwerty even i don understand what is asked in ques 3
11#14 q1
2MnO4- + 5C2O42- + 16H+ --------> 2Mn2+ + 10 CO2 + 8 H2O
now meq of KMnO4 = 8.5 x 110 = 0.85
meq of KMnO4 = meq of oxalic acid = 0.85
now mmoles of oxalic acid = 0.852 = 0.425
0.425 mmoles of oxalic acid is present in 10 ml
so in 200 ml mmoles of oxalic acid = 0.425 x 200 / 10 = 8.5 mmoles
so wgt of oxalic acid = 8.5 x 90 = 765 mg = 0.765 g
so % purity = 0.7651.2 x 100 = 63.75 %
11@cute_cat....is the answer i got for the 3rd one correct...i assumed that they have asked the product of the roots...(alpha)(beta)(gamma)
...??
1Which of these has lowest heat of hydrogenation?
1)1-butene
2)trans 2-butene
3)cis 2-butene
4)1,3 butadiene
1if cos x = tan x, cos y = tan z, and cos z = tan x, then a value of sin x is equal to
a 2 cos 18
b cos 18
c sin 18
d 2 sin 18