B.

V_m < 22.4 L

$$\begin{gathered} \mathrm{compressibility} \left(\mathrm{Z}\right) = \frac{\mathrm{PV}}{\mathrm{nRT}} \\ \mathrm{Z} \mathrm{less} \mathrm{than} \mathrm{unity}, \mathrm{therefore} \\ \frac{\mathrm{PV}}{\mathrm{nRT}} <1 \\ \mathrm{or} \mathrm{PV}<\mathrm{nRT} \\ [\mathrm{At} \mathrm{STP}, \mathrm{P} =1 \mathrm{atm}, \mathrm{T} = 273\mathrm{K}, \mathrm{R} =0.0821 \mathrm{L} \mathrm{atm} {\mathrm{K}}^{-1}] \\ \mathrm{i}.\mathrm{e} \\ \mathrm{or} 1 \mathrm{atm} \mathrm{x} \mathrm{V} < 1 \mathrm{x} 0.0821 \mathrm{x} 273 \\ \mathrm{or} \mathrm{V} <\hspace{0.17em}22.4 \mathrm{litres}. \end{gathered}$$

A.

Principal quantum number = n
Azimuthal quantum number = l = 0 to (n-1)
Magnetic quantum number = m = -l to +1
Spin quantum number = s = +1/2 or -1/2
Now,

(i) In first option the given values are,

n=3; l=2; m=-3; s=-1/2

So according to Azimuthal quantum number

l = 0 to (n-1) = 0 to 2 = here it is 2 which is permissible

m = -l to +l = -2 to +2 = here it is 3 which is  not permissible

s = +1/2 or -1/2 = here it is -1/2 whch is permissible

(ii) In the second option given values are,

n = 4; l = 0, m = 0; s= 1/2

So according to Azimuthal quantum number

l = 0 to (n-1) = 0 to 3 = here it is 0 which is permissible

m =-l to +l = -3  to +3 = here it is 0 which is permissible

s = +1/2 or -1/2 = here it is 1/2 whch is permissible

(iii) In the third option given values are,

n =5; l = 3; m= 0; s= -1/2

So according to Azimuthal quantum number

l = 0 to (n-1) = 0 to 4 = here it is 3 which is permissible

m = -l to +l = -4 to +4 = here it is 0 which is permissible

s = +1/2 or -1/2 = here it is -1/2 whch is permissible

(iv) it the forth option given values are,

n=3; l =2; m=-2; s = 1/2

So according to Azimuthal quantum number

l = 0 to (n-1) = 0 to 2 = here it is 2 which is permissible

m = -l to +l = -2 to +2 = here it is 2 which is permissible

s = +1/2 or -1/2 = here it is -1/2 whch is permissible

A.

C₂H₆

C₂H₆ is non-polar compound and other compounds are polar in nature. Hence, it is least soluble in water.

A.

H₂CO₃

SiF₄ and HClO₃ have sp³ - hybridisation.

BF₃ and H₂CO₃ have sp² - hybridisation but BF₃ is non-polar.

Therefore, the Correct answer is H₂CO₃ 

D.

-35.5 kJ

$$\begin{gathered} \mathrm{C} +{\mathrm{O}}_2 \stackrel{ }{\to } {\mathrm{CO}}_2; ∆\mathrm{G} = - 490 \mathrm{J} ...\left(\mathrm{i}\right) \\ {\mathrm{H}}_2 + \frac{1}{2}{\mathrm{O}}_2 \stackrel{ }{\to } {\mathrm{H}}_2\mathrm{O} ; ∆\mathrm{H} =- 240 \mathrm{J} ...\left(\mathrm{ii}\right) \\ 8\mathrm{C} + 9{\mathrm{H}}_2 \stackrel{ }{\to } {\mathrm{C}}_8{\mathrm{H}}_{18}; ∆\mathrm{H} = +160 \mathrm{J} .... \left(\mathrm{iii}\right) \end{gathered}$$

On applying, 8 x Eq. (i)  + 9x Eq. (ii) -Eq. (iii), we get

$$\begin{gathered} {\mathrm{C}}_8{\mathrm{H}}_{18} + \frac{25}{2}{\mathrm{O}}_2 \stackrel{ }{\to } 8{\mathrm{CO}}_2 + 9{\mathrm{H}}_2\mathrm{O} \\ ∆{\mathrm{H}}_{\mathrm{R}}^0 = [8 \mathrm{x} (-490\left)\right] + [9 \mathrm{x} (-240\left)\right] + 160 \\ = -5920 \mathrm{J} {\mathrm{mol}}^{-1} \end{gathered}$$

Hence, energy exchange when 6 moles of octane is burnt in air = - 5920 x 6 = -35520 J = -35.5 kJ