Ncert Solutions Chemistry Class 11th
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1.32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
Isotope
Isotopic molar mass
Abundance
36Ar
35.96755g mol-1
0.337%
38Ar
37.96272 g mol-1
0.063%
40Ar
39.9624 g mol-1
99.600%
1.32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
|
Isotope |
Isotopic molar mass |
Abundance |
|
36Ar |
35.96755g mol-1 |
0.337% |
|
38Ar |
37.96272 g mol-1 |
0.063% |
|
40Ar |
39.9624 g mol-1 |
99.600% |
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.
Upvote 
1.31. How many significant figures should be present in the answer of the following calculations?
(i) 
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215
1.31. How many significant figures should be present in the answer of the following calculations?
(i)
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215
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
Upvote 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
Upvote 1.29. Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).
1.29. Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).
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 &n
Upvote 1.28. Which one of the following will have largest number of atoms?
(i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl2(g) (Atomic masses: Au = 197, Na = 23, Li = 7, Cl = 35.5 amu)
1.28. Which one of the following will have largest number of atoms?
(i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl2(g) (Atomic masses: Au = 197, Na = 23, Li = 7, Cl = 35.5 amu)
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
Theref
Upvote 1.27. Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 µs
(iii) 25365 mg
1.27. Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 µs
(iii) 25365 mg
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
Upvote 1.26. If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour could be produced?
1.26. If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour could be produced?
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
Upvote 1.25. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?
1.25. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?
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 t
Upvote 1.24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N2 (g) + 3H2(g) —–> 2NH3 (g)
(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 ×103 g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
1.24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N2 (g) + 3H2(g) —–> 2NH3 (g)
(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 ×103 g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
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 = 24
Upvote 1.23. In the reaction,
A + B2——> AB2,
identify the limiting reagent, if any, in the following reaction mixtures
(i) 300 atoms of A + 200 molecules ofB
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
1.23. In the reaction,
A + B2——> AB2,
identify the limiting reagent, if any, in the following reaction mixtures
(i) 300 atoms of A + 200 molecules ofB
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
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
Upvote