NEET Practice Questions (MCQs) with Answers & Solutions

(c) Moles of ${\mathrm{N}}_3^{-}$ ion = 4.2/42=0.1

Each nitrogen atom has 5 valence electrons. Therefore, total number of electrons of ${\mathrm{N}}_3^{-}$ ion=16

Total number of electrons in 0.1 mole or 

4.2g of ${\mathrm{N}}_3^{-}$ ion= 0.1x16xN_A 

=1.6N_A 

As only 50% of the expected product is formed, hence only 10L of NH₃ is formed.

Thus, for the production of 10L of NH₃ , 5L of  N₂ and 15L of H₂ are used and composition of gaseous mixture under the a fore said condition in the end is 

H₂ = 30-15=15L

N₂ = 30-5=25L

NH3 =10L

(D) Weight of NH₃ = 4.25g

Number of moles of NH₃ = $\frac{Weight}{Molecular weight}=\frac{4.25}{17}$=0.25 mol

Number of molecules in 0.25 mole of NH₃ = 0.25 x 6.023 x 10²³

So, number of atoms = 4 x 0.25 x 6.023 x 10²³

=6.0 x 10²³

(c) 4NH₃(g)+ 5O₂(g) $\to$4NO(g) +6H₂O(l)

      4mol         5 mol       4mol        6mol

According to the equation,

5 moles of O₂ required = 4 moles of NH₃

1 mole of O₂ required =4/5= 0.8 moles of NH₃

While 1mole of NH₃ requires = 5/4=1.25 moles of O₂ 

As there is 1 mole of NH₃ and 1 mole of O₂ , so all the oxygen will be consumed.

(d) Weight of ²⁰⁰X = 0.90 x 200 = 180.00 u

Weight of ¹⁹⁹X =0.08 x 199=15.92 u

Weight of ²⁰²X = 0.02 x 202=4.04 u 

Total weight = 199.96 = 200u

Na₂SO₄.10H₂O

(b) Mg + 2HCl $\to$MgCl₂ + H₂

Therefore, 

12g Mg gives 1/2 mole of H₂

(d) m = $\frac{15}{98\times {\displaystyle \frac{(100 \times 1.1-15)}{1000}}}$=1.6

(c) Meq. of NaH₂PO₄ =Meq. of NaOH;

Thus, 12/120/2*1000 = 1 x V

V = 200 mL