# Example-1

Example-1

One mole of a mixture of N

_{2}, NO_{2}, and N_{2}O_{4}has a mean molar mass of 55.4. On heating to a temperature at which all the N_{2}O_{4}may be presumed to have dissociated: N_{2}O_{4}â‡Œ 2NO_{2}, the mean molar mass tends to the lower value of 39.6. What is the mole ratio of N_{2}: NO_{2}: N_{2}O_{4}in the original mixture?- 0.5 : 0.1 : 0.4
- 0.6 : 0.1 : 0.3
- 0.5 : 0.2 : 0.3
- 0.6 : 0.2 : 0.2

Solution A

Let mole of N

_{2}=*x*, mole of NO_{2}=*y*,Mole of N

_{2}=*z*âˆ´ â€¦(i)

If N

_{2}O_{4}â†’ 2NO_{2},â‡’ â€¦(ii)

By dividing Eq. (i) by Eq. (ii), we get

1 +

*z*=*z*= 0.4 mol

Given

*x*+*y*+*z*= 1Putting the value of

*z*is Eq. (i), we get28

*x*+ 46*y*+ 92 + 0.4 = 55.4â€¦(iii)28

*x*+ 46*y*= 18.6â€¦(iv)By Eqs. (iii) and (iv),

*y*= 0.1

âˆ´

*x*= 0.5,*y*= 0.1,*z*= 0.4# Example-2

Example-2

One mole of iron (Fe) reacts completely with 0.65 mol O

**to give a mixture of only FeO and Fe**_{2}_{2}O_{3}. The mole ratio of ferrous oxide to ferric oxide is- 3 : 2
- 4 : 3
- 20 : 13
- None of these

Solution

Let the mole of Fe undergoing formation of FeO be

*x.*Let the mole of Fe undergoing formation of Fe

_{2}O_{3}be (1 âˆ’*x*).Then

As given, = Total moles of oxygen

*x*= 0.4 moles of FeO

= 0.3 = moles of Fe

_{2}O_{3}â‡’

# Example-3

Example-3

In an organic compound of molar mass greater than 100 containing only C, H, and N, the percentage of C is six times the percentage of H, while the sum of the percentages of C and H is 1.5 times the percentage of N. What is the least molar mass?

- 175
- 140
- 105
- 210

Solution B

Compound has C, H, and N is some ratio of % for which data are given as:

If % of C =

*x*, H =*y*, and N =*z*, thenTherefore, by Eqs. (i) and (ii), we have

6

*y*+*y*= 1.5*z*7

*y*= 1.5*z*14

*y*= 3*z*â‡’ â‡’ 18 : 3 : 14

Mole ratio of

Number mole ratio

Empirical formula is C

_{3}H_{6}N_{2}Empirical formula mass = 12 Ã— 3 + 6 + 14 Ã— 2 = 70

Molecular mass > 100

âˆ´ Value of

*n*=Least value of

*n*which is less than 1 is 2. Therefore,Least molecular weight = Empirical formula weight Ã—

*n*= 70 Ã— 2 = 140

# Example-4

Example-4

About 500 mL of 0.1 M KCl, 200 mL of 0.01 M NaNO

_{3}, and 500 mL of 0.1 M AgNO_{3}was mixed. The molarity of K^{+}, Ag^{+}, Cl^{â€“}, Na^{+}, NO_{3 }^{â€“}in the solution would be- [K
^{+}] = 0.04, [Ag^{+}] = 0.04, [Na^{+}] = 0.002, [Cl^{âˆ’}] = 0.04, [NO_{3}^{âˆ’}] = 0.042 - [K
^{+}] = 0.04, [Na^{+}] = 0.00166, [NO_{3}^{âˆ’}] = 0.04166 - [K
^{+}] = 0.04, [Ag^{+}] = 0.05, [Na^{+}] = 0.0025, [Cl^{âˆ’}] = 0.05, [NO_{3}^{âˆ’}] = 0.0525 - [K
^{+}] = 0.05, [Na^{+}] = 0.0025, [NO_{3}^{âˆ’}] = 0.0525

Solution A

In this question, AgNO

_{3}will react with KCl, and AgCl will get precipitated.Number of moles of KCl = 0.1 Ã— = 0.05

Number of moles of NaNO

_{3}= Ã— 0.01 = 0.002Number of moles of AgNO

_{3}= Ã— 0.1 = 0.05Hence,

Number of moles of AgCl precipitated = 0.05

Hence, in solution, only K

^{+}ions and NO_{3}^{âˆ’}will be present apart from NaNO_{3}, and the number of moles will be 0.05 each.After mixing,

Total volume = 500 + 200 + 500 = 1200 mL

After mixing,

Molarity of K

^{+}and NO_{3}^{âˆ’}ions left apart from NaNO_{3 } each

Number of moles NaNO

_{3}in solutionâ‡’ [Na

^{+}] = 0.00166, [NO_{3}^{âˆ’}] = 0.00166Hence, Total [K

^{+}] = 0.04 M, [Na^{+}] = 0.00166[NO

_{3}^{âˆ’}] = 0.04 + 0.00166 = 0.4166.Hence, option (b) is correct, while (a), (c), and (d) are incorrect.

# Example-5

Example-5

A gaseous mixture contains oxygen and nitrogen in the ratio of 1 : 4 by weight. The ratio of their number of molecules is

- 1 : 4
- 1 : 8
- 7 : 32
- 3 : 16

Solution C

Let mass of oxygen = 1 g

Then mass of nitrogen = 4 g

Molecular weight of N

_{2}= 28 gMolecular weight of O

_{2}= 32 g28 g of N

_{2}has = 6.02 Ã— 10^{23}molecules of nitrogen4 g of N

_{2}has molecules of nitrogen Ã— molecules of nitrogen

32 g of O

_{2}= 6.02 Ã— 10^{23}molecules of oxygenâˆ´ 1 g of O

_{2}Ã— 1 molecules of oxygen

Thus,

Ratio of molecules of oxygen : nitrogen