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NEET chemistry

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Question
CBSEENCH11008267
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pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product Ksp of Ba(OH)2 is

  • 3.3 x 10-7

  • 5.0 x 10-7

  • 4.0 x 10-6

  • 5.0 x 10-6

Solution

B.

5.0 x 10-7

Given, pH of Ba(OH)2 = 12
pOH = 14-pH
= 14-12 = 2
We know that,
pOH = -log [OH-]
2 =-log [OH-]
[OH-] = antilog (-2)
[OH-] = 1 x 10-2
Ba(OH)2dissolves in water as 
stack Ba left parenthesis OH right parenthesis subscript 2 space left parenthesis straight s right parenthesis with straight s space mol space straight L to the power of negative 1 end exponent below space rightwards harpoon over leftwards harpoon space stack Ba to the power of 2 plus end exponent with straight s below space plus stack 2 OH to the power of minus with 2 straight s below left square bracket OH to the power of minus right square bracket space equals space 2 straight s space equals space 1 space straight x space 10 to the power of negative 2 end exponent left square bracket Ba to the power of 2 plus end exponent right square bracket space equals space fraction numerator left square bracket OH to the power of minus right square bracket over denominator 2 end fraction space equals space fraction numerator 1 space straight x space 10 to the power of negative 2 end exponent over denominator 2 end fraction straight K subscript sp space equals space left square bracket Ba to the power of 2 plus end exponent right square bracket left square bracket OH to the power of minus right square bracket squared space equals space open parentheses fraction numerator 1 space straight x space 10 to the power of negative 2 end exponent over denominator 2 end fraction close parentheses left parenthesis 1 straight x 10 to the power of negative 2 end exponent right parenthesis squared space equals space 0.5 space straight x space 10 to the power of negative 6 end exponent space equals space 5 space straight x space 10 to the power of negative 7 end exponent

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Question
CBSEENCH11008269
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Maximum number of electrons in a subshell with l =3 and n=4 is

  • 14

  • 16

  • 10

  • 12

Solution

A.

14

n represents the main energy level and l represents the subshell.
If n=4 and l = 3, the subshell is 4f.
If f subshell, there are 7 orbitals and each orbital can accommodate a maximum number of electrons, so,  maximum number of electrons in 4f subshell = 7 x 2 = 14

Question
CBSEENCH11008270
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In which of the following reactions, standard reaction entropy changes (ΔSo) is positive and standard Gibb's energy change (ΔGo) decreases sharply with increasing temperature?

  • C (graphite) +1/2 O2 (g) → CO (g)

  • CO (g) +1/2 (g) → CO2 (g)

  • Mg(s)  +1/2O2 (g) → MgO (s)

  • 1/2 C (graphite) +1/2 O2 (g) → 1/2 CO2 (g)

Solution

A.

C (graphite) +1/2 O2 (g) → CO (g)

Among the given reactions only in te case of 
C (graphite) +1/2 O2 (g) → CO (g)
entropy increases because randomness (disorder) increases. Thus, standard entropy change (ΔSo) is positive.
Moreover, it is a combustion reaction and all the combustion reactions are generally exothermic, ie. ΔHo=-ve
We know that
ΔGo = ΔHo-TΔSo
ΔGo = -ve-T(+ve)
Thus, as the temperature increases, the value of ΔGo decreases.

Question
CBSEENCH11008271
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Equimolar solutions of the following substances were prepared separately, which one of these will record the highest pH value?

  • BaCl2

  • AlCl3

  • LiCl

  • BeCl2

Solution

A.

BaCl2

BaCl2 is a salt of strong acid HCl and strong base Ba(OH)2. So, its aqueous solution is neutral with pH 7. All other salts give acidic solution due to cationic hydrolysis, so their pH is less than 7. Thus, pH value is highest for the solution of BaCl2.

Question
CBSEENCH11008272
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50 mL of each gas A and of gas B takes 150 and 200 s respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas B is 36, the molecular mass of gas A will be

  • 96

  • 128

  • 20.2

  • 64

Solution

C.

20.2

Given,
VA =VB = 50 mL
TA = 150 s
TB = 200 s
MB = 36
MA = ?
From Graham's law of effusion
fraction numerator straight r subscript straight B over denominator straight r subscript straight A space end fraction space equals space square root of M subscript A over M subscript B end root space equals space fraction numerator V subscript B T subscript A over denominator T subscript B. V subscript A end subscript end fraction rightwards double arrow space square root of M subscript A over 36 end root space equals space fraction numerator V subscript A x 150 over denominator 200 space x space V subscript A end fraction O r space square root of M subscript A over 36 end root space equals space 15 over 20 space equals space 3 over 4 M subscript A over 36 space equals space 9 over 16 M subscript A space equals space fraction numerator 9 space x space 36 over denominator 16 end fraction space equals space fraction numerator 9 space x space 9 space over denominator 4 end fraction space equals space 81 over 4 space equals space 20.2 space

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