Ncert Solutions Chemistry Class 11th

Let x be the oxidation number to the underlined elements in the given species:

(a) NaH₂PO₄

(+1) + 2 (+1) + x + 4 (-2) = 0

x + 3 – 8 = 0

x = +5

(b) NaHSO₄

(+1) + (+1) + x + 4 (-2) = 0

x – 6 = 0

x = +6

(c) H₄P₂O₇

4 (+1) + 2x + 7 (-2) = 0

2x -10 =0

x = +5

(d) K₂MnO₄

2 (+1) + x + 4 (-2) = 0

x – 6 = 0

x = +6

(e) CaO₂

2 + 2x = 0

x = -1

(f) NaBH₄

1 + x + 4 (-1) = 0 (Since H is present as hydride ion.)

x = +3

(g) H₂S₂O₇

2 (+1) + 2x + 7 (-2) = 0

x = +6

(h) KAl (SO₄)₂.12H₂O

+1 + 3 + 2x + 8 (-2) + 12 (2 x 1 - 2) = 0

x = +6

5.18. Given, P₁ = P₂ and V₁ = V₂

We know that P₁V₁ = P₂V₂

Or,                n₁RT₁ = n₂RT₂

i.e.,                  n₁T₁ = n₂T₂

 Substituting n = w/M, we get

(W₁/M₁) x T₁ = (W₂/M₂) x T₂

(2.9/M₁) x (95 + 273) = (0.184/2) x (17 + 273)

M₁ = (2.9 x 368 x 2) / (0.184 x 290) = 40 g mol^-1

5.17. No. of moles of CO₂ = Given mass of CO₂ / Molar mass

                                           = 8.8g / 44g mol^-1 = 0.2 mol

Pressure of CO₂ = 1 bar

RR = 0.083 bar dm³ K^–1 mol^–1

T = 273 + 31.1 K = 304.1 K

According to ideal gas equation,

PV = nRT

Therefore, V = nRT/P

                      = (0.2 x 0.083 x 304.1) / 1 bar = 5.048 L

5.15. Molar mass of O₂ = 32 g/mol

It means, 8 g of O₂ has 8/32 mol = 0.25 mol

Molar mass of H₂ = 2 g/mol

It means, 4 g of H₂ has 4/2 mol = 2 mol

Therefore, total number of moles, n = 2 + 0.25 = 2.25 mol

Given, V = 1dm³, T = 27°C = 300 K, R = 0.083 bar dm³ K^-1 mol^-1

Applying PV = nRT,

P = nRT / V

   = (2.25) (0.083 bar dm³ K^-1 mol^-1) (300 K) / (1dm³)

   = 56.025 bar

5.14. Time taken to distribute 10¹⁰ wheat grains = 1s

Time taken to distribute Avogadro number of wheat grains = (1s x 6.022 x 10²³) / 10¹⁰

= 6.022 x 10¹³ s

= (6.022 x 10¹³ / 60 x 60 x 24 x 365) year

= 1.9 x 10⁶ years

5.13. Molecular mass of N₂ = 28g
28 g of N₂ has No. of molecules = 6.022 x 10²³ 

1.4 g of N₂ has No. of molecules = (6.022 x 10²³ x 1.4 g)/28 g= 3.011 x 10²² molecules.
Atomic No. of Nitrogen (N) = 7
1 molecule of N₂ has electrons = 7 x 2 = 14
3.011 x 10²² molecules of N₂ have electrons= 14 x 3.011 x 10²²= 4.215 x10²³ electrons.

5.12. Given,

P= 3.32 bar

V= 5 dm³ 

n= 4 mol

R= 0.083 bar dm³ K^-1 mol^-1

PV = nRT

Or T = PV / nR = 3.32 x 5 dm³ / (4.0 mol x 0.083 bar dm³ K^-1 mol^-1)

         = 50 K

9.56. Dihydrogen is prepared from water by the action of alkali metals like Na and K which is a strong reducing agent.

2Na + 2H₂O → 2NaOH + H₂

2K + 2H₂O → 2KOH + H₂