Chemistry

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

1.22. Speed = Distance / Time

Or, Distance = Speed x Time = 3.0 * 10⁸ms^–1 x 2.00 ns = 3.0 * 10⁸ms^–1 x 2.00 x 10^-9 s = 6 x 10^-1 m = 0.600 m