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Q. No.The empirical formula and molecular mass of a compound are and 180 g, respectively. The molecular formula of the compound is:
1. C9H18O9
2. CH2O
3. C6H12O6
4. C2H4O2
The molar mass of naturally occurring Argon isotopes is:
| Isotope | Isotopic molar mass | Abundance |
| 36-Ar | 35.96755 g mol–1 | 0.337% |
| 38-Ar | 37.96272 g mol–1 | 0.063% |
| 40-Ar | 39.9624 g mol–1 | 99.600% |
| 1. | 49.99947 g mol-1 | 2. | 39.99947 g mol-1 |
| 3. | 35.59947 g mol-1 | 4. | 45.59947 g mol-1 |
On complete combustion, 44 g of a sample of a compound gives 88 g CO2 and 36 g of H2O. The molecular formula of the compound may be:
1.
2.
3.
4.
An organic compound contains carbon, hydrogen, and oxygen. Its elemental analysis gave C, 38.71%, and H, 9.67%. The empirical formula of the compound would be:
| 1. | CH3O | 2. | CH2O |
| 3. | CHO | 4. | CH4O |
| Assertion (A): | The empirical mass of ethene is half of its molecular mass. |
| Reason (R): | The empirical formula represents the simplest whole-number ratio of the various atoms present in a compound. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | (A) is False but (R) is True. |
1. Fe3O2
2. Fe2O2
3. Fe2O3
4. Fe3O4
If 2.74 g of the metal oxide contains 1.53 g of metal, then the empirical formula of vanadium oxide is:
(Atomic Mass of V = 52)
1.
2.
3.
4.
The hydrated salt, Na2SO4.nH2O, undergoes a 55% loss in mass on heating and becomes anhydrous. The value of n will be:
| 1. | 5 | 2. | 3 |
| 3. | 7 | 4. | 10 |
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Q. No.
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