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प्रश्न
The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temperature and pressure.
- Which sample of gas contains the maximum number of molecules?
- If the temperature and pressure of gas A are kept constant, then what will happen to the volume of A when the number of molecules is doubled?
- If this ratio of gas volume refers to the reactants and products of a reaction, which gas law is being observed?
- If the volume of A is actually 5.6 dm3 at STP, calculate the number of molecules in the actual Volume of D at STP (Avogadro's number is 6 × 1023).
- Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N2O).
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उत्तर
(i) D contains the maximum number of molecules because the volume is directly proportional to the number of molecules.
(ii) The volume will double because it is directly proportional to the number of molecules at constant temperature and pressure.
`("V"_1)/("V"_2)` = `("n"_1)/("n"_2)`
`("V"_1)/("V"_2)` = `("n"_1)/(2"n"_1)`
So, V2 = 2V1
(iii) Gay Lussac's law of combining volume is being observed.
(iv) The volume of D = 5.6 × 4 = 22.4 dm3, so the number of molecules = 6 x 1023 because according to mole concept 22.4 litre volume at STP has 6 x 1023 molecules.
(v) No. of moles of D = 1 because volume is 22.4 litre.
∴ Mass of N2O = (14 × 2 + 16 × 1) = 44 g
संबंधित प्रश्न
How does Avogadro's law explain Gay - lussac's law of combining volumes?
When gases react their volumes bear a simple ratio to each other, under the same conditions of temperature and pressure. Who proposed this gas law?
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\[\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}\]
What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen?
24 cc Marsh gas (CH4) was mixed with 106 cc oxygen and then exploded. On cooling the volume of the mixture became 82 cc, of which, 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculations.
What volume of oxygen would be required to burn completely 400 ml of acetylene [C2H2]? Also calculate the volume of carbon dioxide formed.
\[\ce{2C2H2 + 5H2O -> 4CO2 + 2H2O(l)}\]
1250 cc of oxygen was burnt with 300cc of ethane [C2H6]. Calculate the volume of carbon dioxide formed:
\[\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}\]
A mixture of hydrogen and chlorine occupying 36 cm3 was exploded. On shaking it with water, 4 cm3 of hydrogen was left behind. Find the composition of the mixture.
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\[\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}\]
Which amongst the following options is the correct graphical representation of Boyle’s law?
