Given that, volume = 500 mL = 0.5 litre

weight of solute (w) = 4.9 g

equivalent weight of

H₃PO₄ = $\frac{\mathrm{molecular} \mathrm{weight}}{\mathrm{basicity}}$

= $\frac{3 \times 1 + 31 + 4 \times 16}{3}$ = $\frac{98}{3}$

Hence, normality of solution (N)

= $\frac{\mathrm{w}}{\mathrm{equivalent} \mathrm{weight} \mathrm{x} \mathrm{volume}}$

= $\frac{4.9}{{\displaystyle \frac{98}{3}\times 0.5}}$ = 0.3N

According to Bohr's atomic model, electrons can revolve only in those orbits, where its angular momentum is a multiple of h/2$\mathrm{\pi }$ ie equal to nh/2$\mathrm{\pi }$ (where n is an integer)

$\mathrm{\alpha }$ - particle can be used to convert ₁₃Al²⁷ into ₁₅P³⁰ .

₁₃Al²⁷ + ₂He⁴ $\to$ ₁₅P³⁰ + ₀n¹

From Einstein equation, E = mc²

$\therefore$ Energy equivalent to 1 amu

(E)= 1 a.m.u. x (3 x 10⁸ m/s)²

= 1.66 x 10^-27 Kg x 9 x 10¹⁶ m²/s²

= 14.94 x 10^-11 joule

(Kg m²/s² = joule)

= $\frac{14.94 \mathrm{x} {10}^{-11}}{1.6 \mathrm{x} {10}^{-13}}\mathrm{MeV}$

= 931.1 MeV

FeCl₃ is a salt of strong acid and weak base, hence on hydrolysis, it gives acidic solution. Hence, the pH value of solution will be less than 7

Given, K_a = 1.8 x 10^-5

[salt] = 1 x 10 = 10 millimol

[Acid] = 2 x 50 = 100 millimol

From Henderson's equation,

pH = pK_a + log $\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$

= - log (1.8 x 10^-5) + log $\frac{10}{100}$

= - log 1.8 + 5 + log 10^-1

= - 0.2553 + 5 - 1 = 3.7447

= 4

For the solution of alcohol and water, A ... B attractions, are lesser than B...B and A ... A attractions, hence the vapour pressure of solution is greater than expected. i.e., this solution shows positive deviation from Raoult's law ($∆$V_m= +ve, $∆$H_m= +ve).

The atomic number of radioactive element increases by one by the emission of 1 $\mathrm{\beta }$ - particle. Hence, after the emission of 2 $\mathrm{\beta }$ -particles, there will be an increase of 2 units in atomic number.

Hence, ₅X¹⁴ $\stackrel{-2\mathrm{\beta } -\mathrm{particle}}{\to }$ ₇N¹⁴

Hence, the number of neutrons in parent element (X) = 14 - 5 = 9

Solubility product (K_sp) of CuS = 10^-31

$\therefore$ Solubility of Cus = $\sqrt{{\mathrm{K}}_{\mathrm{sp}}}$ = $\sqrt{{10}^{-31}}$

= 3.16 x 10^-16 mol/lit

Similarly, solubility of Ag₂S

= $3\sqrt{\frac{{\mathrm{K}}_{\mathrm{sp}}}{4}}$

($\because$ for Ag₂S, 4s³ = K_sp)

= $\sqrt{\frac{{10}^{-42}}{4}}$

= 6.3 x 10^-15 mol/litre

Solubility of HgS = $\sqrt{{\mathrm{K}}_{\mathrm{sp}}}$

= $\sqrt{{10}^{-54}}$

= 10^-27 mol/lit

Hence, the correct order of solubility is

Ag₂S > Cus > HgS

We know that,

Molality = $\frac{\mathrm{no}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{weight} \mathrm{of} \mathrm{solvent} (\mathrm{in} \mathrm{kg})}$

$\therefore$ 2.0 = $\frac{0.5}{\mathrm{weight} \mathrm{of} \mathrm{solvent} (\mathrm{in} \mathrm{kg})}$

$\therefore$ weight of solvent = $\frac{0.5}{2.0}$ = 0.250 Kg = 250 g