NEET Practice Questions (MCQs) with Answers & Solutions

Molar mass of 1L of gas = mass of 1L N₂

∴ Molecular masses will be equal i.e., molecular mass of the gas = 28, hence formula is C₂H₄

∵ 6.02 × 10²³ atoms of C = 1 mole of C

∴ 2.65 × 10²² atoms of C = $\frac{1\times 2.65\times {10}^{22}}{6.02\times {10}^{23}}$ mole = $\frac{2.65}{6.02\times 10}=0.044$ mole

Now,

Thus, the empirical formula of the compound is Na₂CO₃.

Calculation of empirical formula

Thus, the empirical formula is CH₄N.

∴ Empirical formula is C₂H₃O₂

∴ Empirical formula mass of C₂H₃O₂= 59

Also, Molecular mass = 2 × Vapour density = 2 × 59 = 118

∴ $n=\frac{\text{Molecular mass}}{\text{Empirical formula mass}}=\frac{118}{59}=2$

Now, Molecular formula = n × (Empirical formula) = 2 × (C₂H₃O₂) = C₄H₆O₄

∴ Molecular formula is C₄H₆O₄. 

The balanced chemical equation for the reaction can be written as:

${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}5/2\hspace{0.17em}\hspace{0.17em}O}}_{\text{2}}\stackrel{}{\to }{\text{\hspace{0.17em}2\hspace{0.17em}CO}}_{\text{2}}{\text{\hspace{0.17em}+\hspace{0.17em}H}}_{\text{2}}\text{O}$

$1Vol. 5/2Vol.$

$1ml5/2ml$

$30ml30\times 5/2=75ml$

Hence, volume of the oxygen at STP required to burn 30 ml of acetylene at STP = 75 ml.  

According to Avogadro's hypothesis, volume occupied by one mole of any gas at STP is 22.4 litres.

$C{H_4}_{\left(g\right)}+2{O_2}_{\left(g\right)}\stackrel{}{\to }C{O_2}_{\left(g\right)}+2H_2{O}_{\left(l\right)}$

           1 mole               2 moles

         2 moles               4 moles

$2\times 16gm=32gm$       $4\times 22.4\text{\hspace{0.17em}\hspace{0.17em}litres\hspace{0.17em}}=\text{\hspace{0.17em}89.6 litres}$

Valency of metal in Z₂O₃ = 3

$Z_2{O}_3+3H_2\stackrel{}{\to }2Z+3H_{2}O$

0.1596 gm of Z₂O₃ react with H₂ = 6 mg = 0.006 gm

1 gm of H₂ react with Z₂O₃ $=\frac{0.1596}{0.006}=26.6gm$

Equivalent wt. of Z₂O₃ = 26.6 = equivalent wt. of Z + equivalent wt. of O = E + 8 = 26.6 or E = 18.6

Valency of metal in Z₂O₃.=3

Equivalent weight = Atomic weight/Valency

Atomic weight of Z = 18.6 × 3 = 55.8  

Let the mass of H₂ gas be x g and mass of O₂ gas 4x g

Molar    H₂:O₂

mass     2:32

i.e         1:16

.^.. Molar ratio = ${\mathrm{n}}_{{\mathrm{H}}_2}/{\mathrm{n}}_{{\mathrm{O}}_2}$ = x/2/4x/32 = x x 32/2 x 4x =4/1 = 4:1

(b) If Avogadro number N_A, is changed from 6.022 x 10²³ mol^-1 to 6.022 x 10²⁰ mol^-1, this would change the mass of one mole of carbon.

._.. 1 mole of carbon has mass = 12 g

or 6.022 x 10²³ atoms of carbon have mass = 12 g

.^.. 6.022 x 10²⁰ atoms of carbon have mass

               = $\frac{12}{6.022 \mathrm{x} {10}^{23}}$ x 6.022 x 10²⁰ = 0.012 g

(d) Key Concept In the given problem we have provided practicalyield of MgO. For calculation of percentage yield of MgO, we need theoretical yield of MgO. For this we shall use mole concept.

              MgCO₃ (s) $\stackrel{ }{\to }$ MgO(s) + CO₂ (g)               ...(i)

$$\begin{gathered} \mathrm{Moles} \mathrm{of} {\mathrm{MgCO}}_3 = \frac{\mathrm{Weight} \mathrm{in} \mathrm{gram}}{\mathrm{Molecular} \mathrm{weight}} \\ = \frac{20}{84} = 0.238 \mathrm{mol} \\ \mathrm{From} \mathrm{Eq}. \left(\mathrm{i}\right) \\ 1 \mathrm{mole} \mathrm{of} \mathrm{MgCO}3 \mathrm{gives} = 1\mathrm{mol} \mathrm{MgO} \\ \therefore 0.238 \mathrm{mole} {\mathrm{MgCO}}_3 \mathrm{will} \mathrm{give} = 0.238 \mathrm{mol} \mathrm{MgO} \\ = 0.238 \mathrm{X} 40\mathrm{g} =9.52\mathrm{g} \mathrm{MgO} \\ \mathrm{Now}, \mathrm{practical} \mathrm{yield} \mathrm{of} \mathrm{MgO} = 8 \mathrm{g} \\ \therefore \% \mathrm{purity} =\frac{8}{9.52} \mathrm{X} 100 = 84\% \\ \mathrm{Alternate} \mathrm{method} \\ {\mathrm{MgCO}}_3 \stackrel{ }{\to }\mathrm{MgO} + {\mathrm{CO}}_2 \\ \therefore 8\mathrm{g} \mathrm{MgO} \mathrm{will} \mathrm{be} \mathrm{form} \mathrm{from} \frac{84}{5}\mathrm{g} \\ \therefore \% \mathrm{purity} = \frac{84}{5}\mathrm{X}\frac{100}{20} = 84\% \end{gathered}$$