What would be the heat released when an aqueous solution containing 0.5 mole of HNO3 is mixed with 0.3 mole of OH-? (enthalpy of neutralisation is- 57.1 kJ)
28.5 kJ
17.1 kJ
45.7 kJ
1.7 kJ
B.
17.1 kJ
0.5 mole of HNO3 equivalents to 0.5 mole of H+. But only 0.3 mole of OH- ions are available.
Thus, 0.3 mole of H+ ions will combine with 0.3 mole of OH- ions to form 0.3 mole of H2O and 0.2 mole of H+ ions will remain unreacted.
Heat evolved when 1 mole of H+ ions combine with 1 mole of OH- ions = 57.1 kJ.
Therefore, heat evolved when 0.3 mole of H+ ions combine with 0.3 mole of OH- ions = 0.3x 57.1 = 17.13 kJ.
The Vividh Bharti Station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (Kilohertz). Calculate the wavelength () of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?
319.4 m and X-rays
319.4 m and radiowave
219.3 m and microwave
219.3 m and radiowave
D.
219.3 m and radiowave
In the above question,
Frequency = v = 1368 103 Hz
= = = 219.3 m
(where, = wavelength; c = velocity of light; v = frequency)
The range of wavelength of radiowaves is 100 m to radiation is 10000 m.
The charge/size ratio of a cation determines its polarising power. Which one of the following sequences represents the increasing order of the polarising power of the cationic species, K+, Ca2+, Mg2+, Be2+?
K+ < Ca2+ < Mg2+ < Be2+
Be2+ < K2+ < Ca2+ < Mg2+
Mg2+< Be2+ < K+ < Ca2+
Ca2+ < Mg2+ < Be+ < K+
A.
K+ < Ca2+ < Mg2+ < Be2+
High charge and small size of the cations increases polarisation. As the size of the given cations decreases as
K+ > Ca2+ > Mg2+ > Be2+
Thus, polarising power increases as
K+ < Ca2+ < Mg2+ < Be2+
Which of the following is correctly arranged in order of increasing weight?
0.0105 equivalent of H2CO2 ·2H2O < 0.625 g of Fe < 0.006 g atom of Ag <6.0 x 1021 atoms of Zn
0.625 g of Fe < 0.0105 equivalent of H2C2O4 ·2H2O < 6.0 x 1021 atoms of Zn < 0.006 g atom of Ag
0.625 g of Fe < 6.0 x 1021 atoms of Zn < 0.006 g atom of Ag < 0.0105 equivalent of H2C2O4 ·2H2O
0.0105 equivalent of H2C2O4. 2H2O < 0.006 g atom of Ag <6.0 x 1021 atoms of Zn < 0.625 g of Fe
C.
0.625 g of Fe < 6.0 x 1021 atoms of Zn < 0.006 g atom of Ag < 0.0105 equivalent of H2C2O4 ·2H2O
0.625 g Fe
0.0105 eq of H2C2O4. 2H2O = 0.0105 x 63
= 0.66 g (Eq. wt. = 63)
6 x 1021 atoms of Zn =
0.006 g atom of Ag= 0.006 x 108= 0.648 g
Therefore, Increasing order of weight
Fe < Zn < Ag < H2CO3 ·2H2O
In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainty with which the position of the electron can be located is
(Given, h= 6.6 x 10-34 kg m2 s-1, mass of electron em = 9.1 x 10-31 kg)
215 x 10-3 m
2.78 x 10-3 m
1.92 x 10-3 m
3.24 x 10-3 m
C.
1.92 x 10-3 m
Given that velocity of electron, v = 600 m/s
Accuracy of velocity = 0.005%
v = = 0.03
According to Heisenberg's uncertainty principle
1.92 10-3 m
What percentage of -D-(+-glucopyranose is found at equilibrium in the aqueous solution?
64%
36%
100%
50%
A.
64%
Ordinary glucose is -glucose, with a fresh aqueous solution has specific rotation, []D = + 110°. On keeping the solution for sometimes, -glucose slowly changes into an equilibrium mixture of -glucose (36%) and -glucose (64%) and the mixture has specific rotation+ 52.5°.
The number average molar mass and mass average molar mass of a polymer are respectively 30,000 and 40,000. The
polydispersity index (PDI) of the polymer is
-1
0
> 1
< 1
C.
> 1
Polydispersity Index or PDI =
PDI =
Therefore, polydispersity index or pDI of the polymer is > 1.
0.40 g of helium in a bulb at a temperature of T K had a pressure of p atm. When the bulb was immersed in water bath at temperature 50 K more than the first one, 0.08 g of gas had to be removed to restore the original pressure. T is
500 K
400 K
600 K
200 K
D.
200 K
As pressure and volume remains constant
n1T1 = n2T2
n1 = = 0.1; T1 = TK
n2 = ; T2 = (T + 50)K
On putting the values,
0.1 x T = 0.08 x (T + 50)
0.1T = 0.08 T = 4
0.1 T - 0.08 T = 4
0.02 T = 4
T = 200K
In an amino acid, the carboxyl group ionises at pKa = 2.34 and ammonium ion at pKa2 = 9.60. The isoelectric point of the amino acid is at pH
4.32
3.34
9.46
5.97
D.
5.97
Isoelectric point, pH =
=
= 5.97
For preparing 3.00 L of 1M NaOH by mixing portions of two stock solutions (A and B) of 2.50 M NaOH and 0.40 M NaOH respectively. Find out the amount of B stock solution (in L) added.
8.57 L
2.14 L
1.28 L
7.51 L
B.
2.14 L
Moles needed = 3.00 L 1.00 M = 3 mol.
Let's assume, x L of 2.50 M NaOH is added. Then (3 - x) L of 0.40 M NaOH is added (Given). ..... (i)
The number of moles of solute from the more concentrated solution is 2.50, that becomes less concentrated solution is (0.40) (3.00 - x).
Total number of moles = 3.00
2.50x + 0.40 (3.00 - x) = 3.00
2.50x + 1.2 - 0.40x = 3.00
2.10x + 1.2 = 3.00
2.10x = 1.8
x = 0.857 L of 2.50 M NaOH ... (ii)
Putting value of x from (ii) in (i), is
3 - 0.857 = 2.14 L of 0.40 M NaOH
Therefore, the amount of B stock solution to be added is 2.14 L.
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