Ncert Solutions Chemistry Class 11th

1.32. Molar mass of argon is

= [ (35.96755 * 0.337/100)+ (37.96272 * 0.063/100)+ (39.9624 * 99.60/100)]g mol^-l

= [0.121+0.024+39.802] g mol^-l

= 39.947 g mol^-l

So, the molar mass of argon is 39.947 g/ mol.

1.31. First we need to find the least precise number to find the significant figures.

(i) Least precise number of the calculation  is 0.112

Therefore, number of significant figures in the answer = Number of significant figures in the least precise number, i.e. 3

(ii) Least precise number of calculations = 5.364

Therefore, number of significant figures in the answer will be = Number of significant figures in 5.364 = 4

(iii) 0.0125+0.7864+0.0215

Since the least number of decimal places in each term is four, the number of significant figures in the answer will also be 4.

1.30. 1 mol of 12C atoms = 6.02 x 10²³ atoms = 12 g

Or, 6.02 x 10²³ atoms of 12C have mass = 12 g

Therefore, 1 atom of 12C will have mass = 12 x 1/6.02 x 10²³ = 1.9927 x 10^-23 g

1.29. According to the formula of mole fraction,

X (C₂H₅OH) = 0.040 = n (ethanol) / [n (ethanol) + n (water)]

Now, n (water) = 1000g/18g mol^-1 = 55.55 moles                                          [? Density of water=1kg m^-3]

Therefore, 0.040 = n (ethanol) / [n (ethanol) + 55.55]

i.e., 0.04 x n (ethanol) + 2.222 = n (ethanol)

i.e., 2.222 = [1- 0.04] n (ethanol)

i.e., n (ethanol) = 2.222 / 0.96 = 2.314 M

1.28. The number of atoms in each of the given elements are calculated as below:

(i) 1 g Au = 1 / 197 mol = 1/ 197 x 6.02 x 10²³ atoms

(ii) 1 g Na = 1/ 23 mol = 1/ 23 x 6.02 x 10²³ atoms

(iii) 1 g Li = 1/ 7 mol = 1 / 7 x 6.02 x 10²³ atoms

(iv) 1 g of Cl₂ = 1/ 71 mol = 1/ 71 x 6.02 x 10²³ atoms

Therefore, since the denominator is the smallest in case of Li, 1 g Li has the largest number of atoms

1.27.  (i) 28.7 pm = 28.7 x 10^-12 m = 2.87 x 10^-11 m

(ii) 15.15 µs = 15.15 x 10^-6 s = 1.515 x 10^-5 s

(iii) 25365 mg = 25365 mg x 10^-6 kg = 2.5365 x 10^-2 kg

1.26. H₂ and O₂ react according to the equation

2H₂ (g) + O₂  (g) ——>2H₂O  (g) 

Thus, 2 volumes of H₂ react with 1 volume of O₂ to produce 2 volumes of water vapour. Hence, 10 volumes of H₂ will react completely with 5 volumes of O₂ to produce 10 volumes of water vapour.

1.25. Molar mass of Na₂CO₃= (2 x 23) + 12 + (3 x 16) = 106g mol^-1 

0.50 mol Na₂CO₃ means 0.50 x 106 g = 53 g of Na₂CO₃

0.50 M Na₂CO₃ means 0.50 mol per volume in litre,

i.e., half of 106 g Na₂CO₃ is present in 1 L solution.

i.e.,53 g Na₂CO₃ is present in 1 L of the solution

1.24. According to the given equation, 1 mol of N₂ reacts with 3 mol of H₂.

Or, 28 g of N₂ react with 6 g of H₂.

So, 2000 g of N₂ will react with H₂ = 6/28 x 2000 g = 428.6 g of H₂.

(i) 2 mol of N₂ or 28 g of N₂ produce NH₃ = 2 mol = 34 g

So, 2000 g of N₂ will produce NH₃ = 34/28 x 2000 g = 2428.57 g

(ii) Yes, N₂ is the limiting reagent while H₂ is the excess reagent. So, H₂ will remain unreacted.

(iii) H₂ will remain unreacted. Mass left unreacted = 1000 g – 428.6 g = 571.4 g

1.23.  (i) According to the given reaction, 1 atom of A reacts with 1 molecule of B
Thus, 200 molecules of B will react with 200 atoms of A and 100 atoms of A will be left unreacted. Hence, B is the limiting reagent while A is the excess reagent.

(ii) According to the given reaction, 1 mol of A reacts with 1 mol of B
Thus, 2 mol of A will react with 2 mol of B. Hence, A is the limiting reactant since 1 mol of B is left unreacted.

(iii) No limiting reagent.

(iv) 2.5 mol of B will react with 2.5 mol of A. Hence, B is the limiting reagent.

(v) 2.5 mol of A will react with 2.5 mol of B. Hence, A is the limiting reagent