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1.	How many gram of CaCO3 be decomposed to produce 20 litre of CO2at 750 torr and 27°C.
Answer» It is thus evident that 1 mole of CO2is obtained by decomposition of I mole of CaCO3.

How many gram of CaCO3 be decomposed to produce 20 litre of CO2at 750 torr and 27°C.

Answer»

It is thus evident that 1 mole of CO2is obtained by decomposition of I mole of CaCO3.

Discussion

2.	4 litre C2H4(g) burns in oxygen at 27°C and 1 atm to produce CO2 (g) and H20(g). Calculate the volume of CO2 formed at(a) 27°C and 1 atm,(b) 127°C and 1 atm,(c) 27°C and 2 atm.
Answer» (a) Under same conditions of P and T, volume of gases react in their inole ratio and produce the products in the same molar ratio. Thus, at 27°C and 1atm.

4 litre C2H4(g) burns in oxygen at 27°C and 1 atm to produce CO2 (g) and H20(g). Calculate the volume of CO2 formed at(a) 27°C and 1 atm,(b) 127°C and 1 atm,(c) 27°C and 2 atm.

Answer»

(a) Under same conditions of P and T, volume of gases react in their inole ratio and produce the products in the same molar ratio. Thus, at 27°C and 1atm.

Discussion

3.	A student forgot to add the reaction mixture to the round bottomed open flask at 27°C and put it on the flame. After a lapse of time, he realized his mistake, using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?
Answer» Initial mole of air in flask = n1 Temperature = 21 + 273 = 300 K Final mole of air in flask = n2 Temperature = 477 + 273 = 750 K The volume and pressure remains constant, thus, n1T1= n2T2 n2= n1x 300/750 = 2/5 n1 Thus, air expelled out on heating = n1- n2= n1- (2/5) n1= 3/5 n1 orfraction of air expelled = 3/5.

A student forgot to add the reaction mixture to the round bottomed open flask at 27°C and put it on the flame. After a lapse of time, he realized his mistake, using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?

Answer»

Initial mole of air in flask = n1

Temperature = 21 + 273 = 300 K

Final mole of air in flask = n2

Temperature = 477 + 273 = 750 K

The volume and pressure remains constant, thus, n1T1= n2T2

n2= n1x 300/750 = 2/5 n1

Thus, air expelled out on heating = n1- n2= n1- (2/5) n1= 3/5 n1

orfraction of air expelled = 3/5.

Discussion

4.	Through the two ends of a glass tube of length 200 cmhydrogen chloride gas and ammonia are allowed to enter. At what distance ammonium chloride will first appear?
Answer» Let a cm from HClend white fumes of NH4Clare noticed. From Graham's law

Through the two ends of a glass tube of length 200 cmhydrogen chloride gas and ammonia are allowed to enter. At what distance ammonium chloride will first appear?

Answer»

Let a cm from HClend white fumes of NH4Clare noticed.

From Graham's law

Discussion

5.	For 10 minutes each, at 27°C, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 3 litre capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen. What is the molar mass of the unknown gas?
Answer» For the diffusion of both gases for 10 minutes under identical conditions

For 10 minutes each, at 27°C, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 3 litre capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen. What is the molar mass of the unknown gas?

Answer»

For the diffusion of both gases for 10 minutes under identical conditions

Discussion

6.	6.0 g He and 12.0 g Ne molecules both having average velocity of 4 x 10^2 ms-1 are mixed. Calculate kinetic energy per mole of the mixture.
Answer» For He: For Ne:

6.0 g He and 12.0 g Ne molecules both having average velocity of 4 x 10^2 ms-1 are mixed. Calculate kinetic energy per mole of the mixture.

Answer»

For He:

For Ne:

Discussion