1

### 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.
2

### 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
3

### NEET 2013 (Karnataka)

In an experiment it showed that 10 mL of 0.05 M solution of chloride required 10 mL of 0.1 M solution of AgNO3, which of the following will be the formula of the chloriode (X stands for the symbol of the element other than chlorine)
A
X2Cl2
B
XCl2
C
XCl4
D
X2Cl

## Explanation

Milimoles of solution of chloride = 0.05 $\times$ 10 = 0.5

Millimoles of AgNO3 solution = 10 $\times$ 0.1 = 1

So, the millimoles of AgNO3 are double than the chloride solution.

$\therefore$ XCl2 + 2AgNO3 $\to$ 2AgCl + X(NO3)2
4

### NEET 2013

6.02 $\times$ 1020 molecules of urea present in 100 mL of its solution. The concentration of solution is
A
0.001 M
B
0.1 M
C
0.02 M
D
0.01 M

## Explanation

Moles of urea = ${{6.02 \times {{10}^{20}}} \over {6.02 \times {{10}^{23}}}} = 0.001$

Concentration of solution = ${{0.001} \over {100}} \times 1000$ = 0.01 M