NEET Chemistry Solved Question Paper 2015

In the above question,

Frequency = v = 1368 $\times$10³ Hz

 $\mathrm{\lambda }$ = $\frac{c}{v}$ = $\frac{3 \times {10}^8 \mathrm{m}/\mathrm{s}}{1368 \times {10}^3 {\mathrm{s}}^{-1}}$ = 219.3 m

(where, $\mathrm{\lambda }$ = wavelength; c = velocity of light; v = frequency)

The range of wavelength of radiowaves is 100 m to radiation is 10000 m.

As pressure and volume remains constant

n₁T₁ = n₂T₂

n₁ = $\frac{0.4}{4}$ = 0.1; T₁ = TK

n2 = $\frac{0.40 - 0.08}{4} = 0.08$; T₂ = (T + 50)K

On putting the values,

$\Rightarrow$0.1 x T = 0.08 x (T + 50)

$\Rightarrow$0.1T = 0.08 T = 4

$\Rightarrow$0.1 T - 0.08 T = 4

$\Rightarrow$0.02 T = 4 

$\Rightarrow$T = 200K

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 = $\frac{6 \times {10}^{21}}{6.023 \times {10}^{23}} \times 65 = 0.647 \mathrm{g}$

0.006 g atom of Ag= 0.006 x 108= 0.648 g

Therefore, Increasing order of weight

Fe < Zn < Ag < H₂CO₃ ·2H₂O

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.

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⁺

Isoelectric point, pH = $\frac{{\mathrm{pK}}_{{\mathrm{a}}_1} + {\mathrm{pK}}_{{\mathrm{a}}_2}}{2}$

                              = $\frac{234 + 9.60}{2}$

                               = 5.97

Given that velocity of electron, v = 600 m/s

Accuracy of velocity = 0.005%

$\therefore$ $∆$v = $\frac{600 \times 0.005}{100}$ = 0.03

According to Heisenberg's uncertainty principle

$∆\mathrm{x} . \mathrm{m}∆\mathrm{v} \ge \frac{\mathrm{h}}{4\mathrm{\pi }}$

$∆\mathrm{x} = \frac{6.6 \times {10}^{-34}}{4 \times 3.14 \times 9.1 \times {10}^{-31} \times 0.03}$

$∆\mathrm{x} =$1.92 $\times$ 10^-3 m

Polydispersity Index or PDI = $\frac{\mathrm{Mass} \mathrm{average} \mathrm{molar} \mathrm{mass}}{\mathrm{Number} \mathrm{average} \mathrm{molar} \mathrm{mass}} = \frac{{\mathrm{M}}_{\mathrm{w}}}{{\mathrm{M}}_{\mathrm{n}}}$

PDI = $\frac{40000}{30000} = 1.3 > 1$

Therefore, polydispersity index or pDI of the polymer is > 1.

Moles needed = 3.00 L $\times$ 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

$\therefore$ 2.50x + 0.40 (3.00 - x) = 3.00

  $\Rightarrow$  2.50x + 1.2 - 0.40x = 3.00

  $\Rightarrow$  2.10x + 1.2 = 3.00

  $\Rightarrow$  2.10x = 1.8

  $\Rightarrow$   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.

Ordinary glucose is $\mathrm{\alpha }$-glucose, with a fresh aqueous solution has specific rotation, [$\mathrm{\alpha }$]_D = + 110°. On keeping the solution for sometimes, $\mathrm{\alpha }$-glucose slowly changes into an equilibrium mixture of $\mathrm{\alpha }$-glucose (36%) and $\mathrm{\beta }$-glucose (64%) and the mixture has specific rotation+ 52.5°.