1
MCQ (Single Correct Answer)

NEET 2016 Phase 2

Suppose the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mole of XY2 weights 10 g and 0.05 mole of X3Y2 weighs 9 g, the atomic weights of X and Y are
A
40, 30
B
60, 40
C
20, 30
D
30, 20

Explanation

Assume the atomic weight of element X is x and element Y is y.

For XY2, $$n = {w \over {Mol.\,\,wt.}}$$

$$0.1 = {{10} \over {x + 2y}}$$

$$ \Rightarrow x + 2y = {{10} \over {0.1}} = 100$$    ...(i)

For X3Y2, $$n = {w \over {Mol.\,\,wt.}}$$

$$0.05 = {9 \over {3x + 2y}}$$

$$ \Rightarrow 3x + 2y = {9 \over {0.05}} = 180$$    ...(ii)

On solving equations (i) and (ii),

we get y = 30 g mol-1

Put the value of y in equation. (i)

x + 2(30) = 100

$$ \Rightarrow $$ x = 100 - 60 = 40 g mol-1
2
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
3
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

4
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.

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