1. Xtra Edge February 2012 - Career Point

Sol. (i) Let m be the mass of electron. Then the mass
of mu-meson is 208 m. According to Bohr's
postualte, the angular momentum of mu-meson
should be an integral multiple of h/2π.
e
r
+3e
nh
∴ (208 m) vr =
2π
nh nh
∴ v =
=
2 π×
208mr
416πmr
…(1)
Since mu-meson is moving in a circular path
therefore it needs centripetal force which is
provided by the electrostatic force between the
nucleus and mu-meson.
2
( 208m)
v 1
∴
=
r 4πε
×
3e×
e
2
r
2
0
3e
∴ r =
2
4πε0
× 208mv
Substituting the value of v from (1) we get
r =
2
3e
× 416πmr
× 416πmr
0
2
4πε
× 208n
h
2 2
h ε0
2
n
⇒ r =
624πme …(2)
(ii) The radius of the first orbit of the hydrogen
atom
2
2
ε0h
=
2
πme
…(3)
To find the value of n for which the radius of
the orbit is approximately the same as that of
the first Bohr orbit for hydrogen atom, we
equate
equation (2) and (3)
n
2 2
h ε0
2
624πme =
ε
0
h
2
πme
1 2
(iii) = 208 R × z
λ
2
⇒ n = 624 ≈ 25
⎡ 1 1 ⎤
⎢ –
2 2 ⎥
⎢⎣
n1
n2
⎥⎦
1 7 2
⇒ = 208 × 1.097 × 10 × 3
λ
⇒ λ = 5.478 × 10 –11 m
⎡ 1 1 ⎤
⎢ –
2 2 ⎥
⎣1
3 ⎦
CHEMISTRY
6. An organic compound CxH2yOy was burnt with
twice the amount of oxygen needed for complete
combustion to CO2 and H2O. The hot gases, when
cooled to 0 ºC and 1 atm pressure, measured 2.24
L. The water collected during cooling weighed 0.9
g. The vapour pressure of pure water at 20 ºC is
17.5 mm Hg and is lowered by 0.104 mm Hg when
50 g of the organic compound is dissolved in 1000
g of water. Give the molecular formula of the
organic compound. [IIT-1983]
Sol. The combustion reaction is
CxH2yOy + x O2 → x CO2 + y H2O
To start with, the amount of O2 taken is 2x. Hence,
after the combustion reaction, we will be left with
the following amounts.
Amount of oxygen left unreacted = x
Amount of carbon dioxide = x
Amount of water = y
When this mixture is cooled to 0 ºC and 1 atm, we
will be left with oxygen and carbon dioxide. Hence,
the amount 2x occupies the given volume of 2.24 L
at STP. Hence,
( 2.
24 / 2)
L
Amount x =
= 0.05 mol
− 1
22.
4Lmol
Now, Mass of water collected = 0.9 g
Amount of water collected,
0.
9g
y = = 0.05 mol
− 1
18g
mol
Thus, the empirical formula of the compound is
C0.05H2 × 0.05 O0.05, i.e. CH2O.
Now, according to Raoult's law
∆ p
– = x2
p*
XtraEdge for IIT-JEE 9 FEBRUARY 2012
i.e.
( 50g
/ M)
0.
104 mmHg
17.
5mmHg
−1
( 50g
/ M)
+ ( 1000g
/ 18g
mol )
Solving for M, we get M = 150.5 g mol –1
Number of repeating units of CH2O in the
150.
5
molecular formula =
≈ 5
12 + 2 + 16
Hence, Molecular formula of the compound is
C5H10O5.
7. The oxides of sodium and potassium contained in a
0.5 g sample of feldspar were converted to the
respective chlorides. The weight of the chlorides
thus obtained was 0.1180 g. Subsequent treatment
of the chlorides with silver nitrate gave 0.2451 g of
silver chloride. What is the percentage of Na2O and
K2O in the mixture ? [IIT-1979]
=