1

### 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

### 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

### 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

### 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