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NEET Chemistry Solved Question Paper 2010

Multiple Choice Questions

100 cm³ of a given sample of H₂O₂ gives 1000 cm³ of O₂ at S.T.P. The given sample is

10% H₂O₂
90% H₂O₂
10 volume H₂O₂
100 volume H₂O₂

10 volume H₂O₂

Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts- either 2OH or an H and HO_2.

$\underset{100 ml}{2 H_{2} O_{2}} \to 2 H_{2} O + \underset{1000 mL}{O_{2}}$

or 1ml of H₂O₂ will give 10 ml of O₂ at STP.

In the ground state of Cu²⁺the number of shells occupied, subshells occupied, filled orbitals and unpaired electrons respectively are

4, 8, 15 0
3, 6, 15, 1
3, 6, 14, 0
4, 7, 14, 2

3, 6, 14, 0

Cu⁺= 1s² 2s² 2p6 3s² 3p⁶ 3d⁶

So, Shells occupied =3, sub- shell occupied = 6, filled orbital = 14 and unpaired electron = 0

CaO and NaCl have the same crystal structure and approximately the same ionic radii. If U is the lattice energy ofNaCI, the approximate lattice energy of CaO is

Lattic energy = $- \frac{q_{1} q_{2}}{r^{2}}$ where q₁ and q₂ are charges on ions and r is the distance between them. Since interionic distances in CaO and NaCl are similar, (larger cation has smaller anion and vice versa) therefore, r is almost the same. Therefore, lattice energy depends only on charge. Since the magnitude of charge on Na⁺ and Cl^-ions is unity and that on Ca²⁺ and O^2-ions is 2 each, therefore, the lattice energy of CaO is four times the lattice energy of NaCl, i.e..4U.

Equal weight of CO and CH₄ are mixed together in an empty container at 300 K. The fraction of total pressure exerted by CH₄ is.

$\frac{16}{17}$
$\frac{7}{11}$
$\frac{8}{9}$
$\frac{5}{16}$

$\frac{7}{11}$

Let tje weight of CO = weight of CH₄

$Moles of CO = \frac{a}{28}$

$Moles of {CH}_{4} = \frac{a}{16}$

Total moles = $\frac{a}{28} + \frac{a}{16}$

$x_{{CH}_{4}} = \frac{a}{\frac{16}{\frac{a}{28} + \frac{a}{16}}} = \frac{a}{16} \times \frac{28 \times 16}{44 a} = \frac{14}{22} = \frac{7}{11}$

fraction of pressure exerted by CH₄ = 7/11

Match List I with List II and select the correct answer using the codes given below the lists:

List I	List II
A. $(\frac{δG}{δP})_{T}$	1. $μ_{π}$
B. $(\frac{δG}{δT})_{P}$	2. T
C. $(\frac{δH}{δS})_{P}$	3. -S
D. $(\frac{δT}{δP})_{H}$	4. P
	5. V

A-5, B-1, C-2, D-4
A-5, B-3, C-2, D-4
A-3, B-5, C-2, D-1
A-5, B-3, C-2, D-1

A-5, B-3, C-2, D-1

From thermodynamics, dG = VdP-SdT

At constant T, dT = 0 so that $(\frac{δG}{δP})_{T} = V$

At constant P dP = 0 so that $(\frac{δG}{δT}) = - S$

Also $μ_{JT} = (\frac{δT}{δP})_{H}$

What is the pH of0.01 M glycine solution? For glycine K_a1= 4.5x10^-3 and K_a2=1.7x10^-10 298 K.

3.0
10.0
7.06
8.2

7.06

Glycine H₃N⁺CH2COO^- is more acidic than basic. Instead of K_b value, the second given K_a value is corresponding to the K_b value ( K_a $\times$ K_b = 10^-14). Hence, overall ionization constant,

K= K_a1 $\times$ K_a2

= 4.5 $\times$ 10-3 $\times$ 1.7 $\times 10^{- 3} = 7.65 \times 10^{- 13}$

$H^{+} = \sqrt{KC} = \sqrt{7.65 \times 10^{- 13} \times 0.01} = \sqrt{0.765 \times 10^{- 14}} = 0.87 \times 10^{- 7} M$

pH =-log (0.87 $\times$ 10^-7) = 7.06

Two flasks X and Y have capacity 1L and 2L respectively and each of them contains 1 mole of a gas. The temperatures of the flasks are so adjusted that average speed of molecules in X is twice as those in Y. The pressure in flask X would be