Chemistry MCQs for NEET — Practice Questions with Answers

The reaction between Na₂CO₃ and Ca(OH)₂ produces 2NaOH and CaCO₃. Given 1kg (1000g) of Na₂CO₃, the mass of NaOH produced can be calculated as: (2 × 40g/mol) / (106g/mol) × 1000g = 755g. Therefore, 755g of NaOH can be obtained from the reaction of 1kg of Na₂CO₃ with Ca(OH)₂.

In sodium amalgam, which is an alloy of sodium and mercury, the oxidation state of sodium is zero. This is because sodium loses its valence electron to mercury, forming a stable ionic compound where sodium has an oxidation state of +1 and mercury has an oxidation state of -1.

$ 0.1MH_2SO_4 = 1000mL solution contains 0.1 mol H_2SO_4 $ $ \therefore 100mL solution contains 0.01 mol H_2SO_4 $ $ \therefore mass of H_2SO_4 = 0.01 \times 98 = 0.98g $ $ 2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O $ 2(40)g 98g (?) 0.98g $ = { 80 \times 0.98 \over 98 } = 0.8g $

$ { Wt. of metal oxide \over Wt. of metal Choride } = { Eq. wt. of metal + Eq. wt. of oxide \over Eq. wt. of metal + Eq. wt. of Choride } $ $ \therefore { 3 \over 5 } = { E + 8 \over E + 35.5 } \therefore E = 33.25 $

$ M .F. C_4 H_8O_2 $ $ \therefore Molar mass = 88 $ $ Vapour density = 88 / 2 = 44 $

Equal volumes under identical conditions contain equal no. of molecules

$ N_{26 \over 14 } O_{74 \over 16 } = N_{1.85} O_{4.625 } \Rightarrows N_2 O_5 $