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

Average atomic mass of oxygen $=\frac{90\times 16+10\times 18}{100}=16.20$  

On heating KClO₃ dissociates as:

$2KCl{O}_3\stackrel{\Delta }{\to }2KCl+3O_2$

2 moles 3 × 22.4 L at STP

$\because$ 2 moles of KClO₃ on heating produces = 67. 2 L of O₂ at STP

$\because$ 0.1 mole of KClO₃on heating produces = $\frac{67.2}{2}\times 0.1L$ = 3.36 L of O₂ at STP

1 mole of CCl₄ vapour = 12 + 4 × 35.5 = 154 gm = 22.4 L at S.T.P.

∴ Density $=\frac{154}{22.4}gm{L}^{-1}=6.875gm{L}^{-1}$  

22400 c.c. = 6.02 × 10²³ molecules

1 c.c. of $N_{2}O=\frac{6.02\times {10}^{23}}{22400}$ molecules

$=\frac{3\times 6.02\times {10}^{23}}{22400}$ atoms (Since N₂O has three atoms)

$=\frac{6.02\times {10}^{23}}{22400}\times 22$ electron (Because number of electrons in N₂O are 22) 

1 mole of K₄[Fe(CN)₆] =6 gm atoms of carbon

0.5 mole of K₄[Fe(CN)₆] = 3 gm atoms of carbon

= 3 × 12 = 36 g

$âˆ\mu$ 1 mole of BaCO₃contains 3 moles of oxygen atoms.

∴ $\frac{1}{2}$ mole (0.5) of BaCO₃ contains 1.5 moles of oxygen atoms.

% of oxygen = $\frac{8}{m+8}\times 100=40$ or, $40(m+8)=800$ or, m + 8 = 20or, m = 12

∴ % of chlorine = $\frac{35.5}{m+35.5}\times 100$ = $\frac{35.5}{12+35.5}\times 100$ = 74.7(Where m is the atomic mass of metal) 

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.