1
MCQ (Single Correct Answer)

AIPMT 2015 Cancelled Paper

A mixture of gases contains H2 and O2 gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two gases in the mixture ?
A
16 : 1
B
2 : 1
C
1 : 4
D
4 : 1

Explanation

Ratio of weight of gases = wH2 : wO2 = 1 : 4

$$ \therefore $$ Number of moles of H2 = $$1 \over 2$$

$$ \therefore $$ Number of moles of O2 = $$4 \over 32$$

Ratio of moles of gases

= nH2 : nO2

= $$1 \over 2$$ : $$4 \over 32$$

= $$1 \over 2$$ $$ \times $$ $$32 \over 4$$ = 4 : 1
2
MCQ (Single Correct Answer)

AIPMT 2015

The number of water molecules is maximum in
A
1.8 gram of water
B
18 gram of water
C
18 moles of water
D
18 molecules of water

Explanation

1.8 g of water = $${{6.023 \times {{10}^{23}}} \over {18}} \times 1.8$$

= 6.023 $$ \times $$ 1022 molecules

$$ \therefore $$ 18 g of water = 6.023 $$ \times $$ 1023 molecules = 1 mole of water

$$ \therefore $$ 18 moles of water = 18 $$ \times $$ 6.023 $$ \times $$ 1023 molecules

3
MCQ (Single Correct Answer)

AIPMT 2015

If Avogadro number NA, is changed from 6.022 $$ \times $$ 1023 mol$$-$$1 to 6.022 $$ \times $$ 1020 mol$$-$$1, this would change
A
the mass of one mole of carbon
B
the ratio of chemical species to each other in a balanced equation
C
the ratio of elements to each other in a compound
D
the definition of mass in units of grams.

Explanation

We know mass of 1 mol (6.022 $$ \times $$ 1023) atoms of carbon = 12 g

If Avogadro number is changed to 6.022 $$ \times $$ 1020 then mass of 1 mol of cabon is

= $${{12 \times 6.022 \times {{10}^{20}}} \over {6.022 \times {{10}^{23}}}}$$ = $$12 \times {10^{ - 3}}\,g$$

$$ \therefore $$ It would change the mass of one mole of carbon.
4
MCQ (Single Correct Answer)

AIPMT 2015

What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO3 is mixed with 50 mL of 5.8% NaCl solution ? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5
A
3.5 g
B
7 g
C
14 g
D
28 g

Explanation

50 ml of 16.9% solution of AgNO3

$$\left( {{{16.9} \over {100}} \times 50} \right)$$ = 8.45 g of AgNO3

nmole = $${{8.45g} \over {(107.8 + 14 + 16 \times 3)g/mol}}$$

= $$\left( {{{8.45g} \over {169.8g/mol}}} \right) = 0.0497\,moles$$

50ml of 5.8% solution of NaCl contain

NaCl = $$\left( {{{5.8} \over {100}} \times 50} \right) = 2.9g$$

nNaCl = $${{2.9g} \over {(23 + 35.5)g/mol}}$$

= 0.0495 moles

AgNO3 + NaCl $$ \Rightarrow $$ AgCl + Na$$ \oplus $$ + Cl$$ \ominus $$
1 mole 1 mole 1 mole
0.049 mole 0.049 mole 0.049 mole of AgCl

n = $${w \over M}$$

$$ \Rightarrow $$ w = (nAgCl) $$ \times $$ Molecular mass

= (0.049) $$ \times $$ (107.8 + 35.5) = 7.02 g

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