Some Basic Concepts of Chemistry
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1.34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at S.T.P.) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
1.34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at S.T.P.) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
1.34. (i) 1 mole (44 g) of CO2 will have 12 g carbon.
So, 3.38 g of CO2 will have carbon = 12g/44g * 3.38
= 0.9217 g
18 g of water will have 2 g of hydrogen.
So, 0.690 g of water contain hydrogen = 2g/18g * 0.6902g
= 0.0767 g
Since carbon and hydrogen are the only constituents of the compo
Upvote 
1.33. Calculate the number of atoms in each of the following:
(i) 52 moles of He
(ii) 52 u of He
(iii) 52 g of He
1.33. Calculate the number of atoms in each of the following:
(i) 52 moles of He
(ii) 52 u of He
(iii) 52 g of He
1.33. (i) 52 moles of Ar
Ans:1 mole of Ar = 6.022 * 1023 atoms of Ar
Therefore, 52 mole of Ar = 52 * 6.022 * 1023 atoms of Ar= 3.131 * 1025 atoms of Ar
(ii) 52 u of He
Ans:1 atom of He = 4u of the He
Or, 4 u of He = 1 atom of He
So, 52 u of
Upvote 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