1

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

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

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

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.

NEET