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CBSE Class 12 Chemistry Solved Question Paper 2018

Short Answer Type

CO(g) and H₂ (g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions?

Write the coordination number and oxidation state of Platinum in the complex [Pt(en)₂Cl₂].

Analysis shows that FeO has a non-stoichiometric composition with formula Fe_0.95O. Give reason.

Complete and balance the following chemical equations :

$F e^{2 +} + M n O_{4}^{-} + H^{+} \overset{}{\to}$

Complete and balance the following chemical equation:

$M n O_{4}^{-} + H_{2} O + I^{-} \overset{}{\to}$

For the reaction, $2 N_{2} O_{5} (g) \overset{}{\to} 4 N O_{2} (g) + O_{2} (g)$ the rate of formation of NO₂ (g) is 2.8 x 10^-3 Ms^-1. Calculate the rate of disappearance of N₂O₅ (g).

Among the hydrides of Group-15 elements, which have the

Lowest boiling point?
Maximum basic character?
Highest bond angle?
Maximum reducing character?

8.
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol–1) in 250 g of water.
(K_f of water = 1.86 K kg mol^–1)

We know,

W₁ = 250 g, w₂ = 60 g, mw₂ = 180 g/mol, K_f = 1.86 k kg mol^–1

$∆ T_{f} = k_{f} x m O r ∆ T_{f} = k_{f} x \frac{w_{2} x 1000}{m . w_{2} x w_{1} (g)} = 1.86 x \frac{60 x 1000}{180 x 250} = \frac{1.80 x 600}{18 x 25} = \frac{1116}{450} ∆ T_{f} = 2.48 K ∆ T_{f} = T_{s o l v e n t - T_{s o l u t i o n}} = 2.48 = 273.15 - T_{s o l u t i o n} T_{s o l u t i o n} = 270.67 K$

An element ‘X’ (At. mass = 40 g mol–1) having f.c.c. the structure has a unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (N_A = 6.022 × 1023 mol^–1 )