Some Basic Concepts in Chemistry MCQs for NEET — Chemistry Questions with Answers

∴ 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)

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

Plan: For the calculation of mass of AgCl precipitated, we find mass of AgNO₃ and NaCl in equal volume with the help of mole concept.

16.9% solution of AgNO₃ means 16.9 g AgNO₃ is present in 100 mL solution.

8.45 g AgNO₃ will present in 50 mL solution

Similarly,

5.8 g NaCl is present in 100 mL solution

2.9 g NaCl is present in 50 mL solution
AgNO₃ + NaCl $\to$AgCl + NaNO₃

Initial mole   8.45/169.8     2.9/58.5       0             0

                     = 0.049          = 0.049  

After reaction     0                 0              0.049     0.049

Mass of AgCl precipitated
= 0.049 x 143.5= 7g

Key Concept Mole is the biggest unit measure of number of molecules/atoms/ions.

._.. 1 mole of water contains molecules = 6.02 x 10²³

.^.. 18 moles of water contain molecules = 18 x 6.02 x 10²³ molecules

Now, 1mole of water = 18 g of water = 6.02 x10²³ and 1.8 g of water contains =6.02 x10²³ molecules

(a) 1.00 molal aqueous solution = 1.0 mole in 1000 g water n_solute = 1; W_solute = 1000 g.

n_solvent = 1000/18 =55.56

X_solute = 1/(1+55.56) = 0.0177