1

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
2

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
3

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
4

AIPMT 2014

When 22.4 liters of H2(g) is mixed with 11.2 litres of Cl2(g), each at S.T.P. the moles of HCl(g) formed is equal to
A
1 mol of HCl(g)
B
2 mol of HCl(g)
C
0.5 mol of HCl(g)
D
1.5 mol of HCl(g)

Explanation

1 mole = 22.4 L at S.T.P

nH2 = ${{22.4} \over {22.4}}$ = 1 mole

nCl2 = ${{11.2} \over {22.4}}$ = 0.5 mole

H2 + Cl2 $\Rightarrow$ 2HCl(g)
Initial 1 mol 0.5 mol 0
Final ( 1 - 0.5)
= 0.5 mol
(0.5 - 0.5)
= 0 mol
2 x 0.5
1 mol

Here Cl2 is limiting reagent. So 1 mole of HCl(g) is formed.