1

### 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.
2

### 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
3

### AIPMT 2014

1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel, Which reactant is left in excess and how much ? (At. wt. Mg = 24, O = 16)
A
Mg. 0.16 g
B
O2, 0.16 g
C
Mg, 0.44 g
D
O2 , 0.28 g

## Explanation

nMg = $1 \over 24$ = 0.0416 moles

nO2 = $0.56 \over 32$ = 0.0175 moles

Mg + ${1\over2} O_2$ $\Rightarrow$ MgO
Initial 0.0416 moles 0.0175 moles 0
Final ( 0.0416 - 2 x 0.0175)
= 0.0066 moles
0 2 x 0.0175

$\because$ Mass of Mg left in excess = 0.0066 $\times$ 24 = 0.16 g
4

### AIPMT 2014

Equal masses of H2, O2 and methane have been taken in a container of volume V at temperature 27oC in identical conditions. The ratio of the volumes of gases H2 : O2 : methane would be
A
8 : 16 : 1
B
16 : 8 : 1
C
16 : 1 : 2
D
8 : 1 : 2

## Explanation

According to Avogadro's hypothesis ratio of the volumes of gases will be equal to the ratio of their no. of moles.

So, no of moles = ${{Mass} \over {Mol.\,\,mass}}$

nH2 = ${w \over 2}$; nO2 = ${w \over 32}$; nCH4 = ${w \over 16}$

So, the ratio will be ${w \over 2}$ : ${w \over 32}$ : ${w \over 16}$ or 16 : 1 : 2