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Example-1

Example-1
One mole of a mixture of N2, NO2, and N2O4 has a mean molar mass of 55.4. On heating to a temperature at which all the N2O4 may be presumed to have dissociated: N2O4 ⇌ 2NO2, the mean molar mass tends to the lower value of 39.6. What is the mole ratio of N2 : NO2 : N2O4 in the original mixture?
  1. 0.5 : 0.1 : 0.4
  2. 0.6 : 0.1 : 0.3
  3. 0.5 : 0.2 : 0.3
  4. 0.6 : 0.2 : 0.2
Solution A
Let mole of N2 = x, mole of NO2 = y,
 
Mole of N2 = z
 
∴ Description: 11998.png…(i)
 
If N2O4 → 2NO2,
 
Description: 12008.png
 
⇒ Description: 12015.png…(ii)
 
By dividing Eq. (i) by Eq. (ii), we get
 
1 + z = Description: 12024.png
 
z = 0.4 mol
 
Given x + y + z = 1
 
Putting the value of z is Eq. (i), we get
 
28x + 46y + 92 + 0.4 = 55.4…(iii)
 
28x + 46y = 18.6…(iv)
 
By Eqs. (iii) and (iv),
 
y = 0.1
 
∴ x = 0.5, y = 0.1, z = 0.4
 

Example-2

Example-2
One mole of iron (Fe) reacts completely with 0.65 mol O2 to give a mixture of only FeO and Fe2O3. The mole ratio of ferrous oxide to ferric oxide is
  1. 3 : 2
  2. 4 : 3
  3. 20 : 13
  4. None of these
Solution
Let the mole of Fe undergoing formation of FeO be x.
 
Let the mole of Fe undergoing formation of Fe2O3 be (1 − x).
 
Then
 
As given, Description: 12073.png = Total moles of oxygen
 
x = 0.4 moles of FeO
 
Description: 12080.png = 0.3 = moles of Fe2O3
 
⇒ Description: 12087.png
 

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?
  1. 175
  2. 140
  3. 105
  4. 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 , then
Description: 12120.png
 
Therefore, by Eqs. (i) and (ii), we have
 
6y + y = 1.5z
 
7y = 1.5z
 
14y = 3z
 
Description: 12129.png
 
⇒ Description: 12137.png ⇒ 18 : 3 : 14
 
Mole ratio of
 
Number mole ratio
 
Empirical formula is C3H6N2
 
Empirical formula mass = 12 × 3 + 6 + 14 × 2 = 70
 
Molecular mass > 100
 
∴ Value of n = Description: 12188.png
 
Least value of n which is less than 1 is 2. Therefore,
 
Least molecular weight = Empirical formula weight × n
= 70 × 2 = 140
 

Example-4

Example-4
About 500 mL of 0.1 M KCl, 200 mL of 0.01 M NaNO3, and 500 mL of 0.1 M AgNO3 was mixed. The molarity of K+, Ag+, Cl, Na+, NO3  in the solution would be
  1. [K+] = 0.04, [Ag+] = 0.04, [Na+] = 0.002, [Cl] = 0.04, [NO3] = 0.042
  2. [K+] = 0.04, [Na+] = 0.00166, [NO3] = 0.04166
  3. [K+] = 0.04, [Ag+] = 0.05, [Na+] = 0.0025, [Cl] = 0.05, [NO3] = 0.0525
  4. [K+] = 0.05, [Na+] = 0.0025, [NO3] = 0.0525
Solution A
In this question, AgNO3 will react with KCl, and AgCl will get precipitated.
 
Number of moles of KCl = 0.1 × Description: 12198.png = 0.05
 
Number of moles of NaNO3 = Description: 12205.png × 0.01 = 0.002
 
Number of moles of AgNO3 = Description: 12213.png × 0.1 = 0.05
 
Hence,
 
Number of moles of AgCl precipitated = 0.05
 
Hence, in solution, only K+ ions and NO3 will be present apart from NaNO3, 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 NO3 ions left apart from NaNO3 Description: 12220.png
Description: 13480.png each
 
Number of moles NaNO3 in solution Description: 12230.png
Description: 12240.png
⇒ [Na+] = 0.00166, [NO3] = 0.00166
 
Hence, Total [K+] = 0.04 M, [Na+] = 0.00166
 
[NO3] = 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. 1 : 4
  2. 1 : 8
  3. 7 : 32
  4. 3 : 16
Solution C
Let mass of oxygen = 1 g
 
Then mass of nitrogen = 4 g
 
Molecular weight of N2 = 28 g
 
Molecular weight of O2 = 32 g
 
28 g of N2 has = 6.02 × 1023 molecules of nitrogen
 
4 g of N2 has Description: 12250.png molecules of nitrogen
 
Description: 12258.png × molecules of nitrogen
 
32 g of O2 = 6.02 × 1023 molecules of oxygen
 
∴ 1 g of O2 Description: 12268.png × 1
Description: 12284.png molecules of oxygen
 
Thus,
Ratio of molecules of oxygen : nitrogen Description: 12291.png
 




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