Question - 11 : - A student forgot to add the reaction mixture to theround bottomed flask at 27°C but put it on the flame. After a lapse of time, herealised his mistake. By using a pyrometer, he found that the temperature ofthe flask was 477°C. What fraction of air would have been expelled out ?

Answer - 11 : -

Sincethe student was working in the laboratory, there is no change in pressure.Thus, Charles’ Law is applicable.
According to given data : V₁ =V L (say) V₂ = ?
T₁ = 27+273 =300 K; T₂ = 477 + 273 =750K
T1 = T2 orV₂ = T1 =\frac { (VL)\times (750K) }{ 300K } = 2.5 VL
Thus, volume of air expelled = 2.5 V – V = 1.5 V 1.5V
Fraction of air expelled = 1.5V = 3 .

Question - 12 : - Calculate the temperature of 4.0 moles of a gasoccupying 5 dm³ at 3.32 bar (R =0.083 bar dm³ K^-1 mol^-1).

Answer - 12 : - According to idea gas equation, PV = n RT or T =[/latex] = PVnR
According toavailable data :
No. of moles of the gas (n) = 4.0 moles
Volume of the gas (V) = 5 dm³
Pressure of the gas (P) = 3.32 bar
Gas constant (R) = 0.083 bar dm³ K^-1 mol^-1

Question - 13 : - Calculate the total number of electrons present in1.4 g of nitrogen gas.

Answer - 13 : -

Molecularmass of N₂ = 28g
28g of N₂ representmolecules = 6.022 × 10²³
1.4 g of N₂ representmolecules = 6.022 × 10²³ × 1.4g =3.011 × 10²²
Atomic number of nitrogen (N) = 7
1 molecule of N₂ haselectrones = 7 × 2 = 14
3.011 × 10²² molecules of N₂ have electrons = 4 ×3.011 × 10²² = 4.215 × 10²³ electrons

Question - 14 : - How much time would it take to distribute oneAvogadro number of wheat grains if 10¹⁰ grainsare distributed each second ?

Answer - 14 : - Time taken to distribute 10¹⁰ grains= 1 s

Question - 15 : - Calculate the total pressure of a mixture of 8 g ofoxygen and 4 g of hydrogen confined in a vessel of 1 dm3 at 27°C (R = 0-083 bardm³K^-1mol^-1)

Answer - 15 : -

Question - 16 : - Pay load is defined as the difference between themass of displaced air and the mass of the balloon. Calculate the pay load whena balloon of radius 10 m and mass 100 kg is filled with helium at 1.66 bar at27°C. (Density of air = 1.2 kg m^-3 andR = 0.083 bar dm³K^-1mol^-1).

Answer - 16 : - Step I Calculation of the mass of displaced air
Radius of balloon (r) = 10 m
Volume of balloon (V) = 4 πr³ = 4 × 22 ×(10 m)³ = 4190.5 m³
Mass of the displaced air = Volume of air (balloon) × Density of air
= (4190.5 m³) × (1.2 kg m^-3) = 5028.6 kg
Step II Calculation of the mass of the filled balloon.

=279.37 × 10³ mol.
Mass of He present = Moles of He × Molar mass of He
= (279.37 × 10³ mol) × (4 g mol^-1)
= 1117.48 × 10³g = 1117.48 kg
Mass of filled balloon = 100 + 1117.48 = 1217.48 kg
Step III. Calcu lation of the pay load
Pay load = Mass of displaced air – Mass of filled balloon
= 5028.6 – 1217.48 =3811.12 kg.

Question - 17 : - Calculate the volume occupied by 8-8 g of Co₂ at 31.1 °C and 1 bar pressure (R = 0.083 bar LK^-1mol^-1).

Answer - 17 : -

Question - 18 : - 2.9 g of a gas at 95°C occupied the same volume as0.184 g of hydrogen at 17°C and at the same pressure. What is the molar mass ofthe gas ?

Answer - 18 : -

Question - 19 : - A mixture of hydrogen and oxygen at one bar pressurecontains 20% by weight of hydrogen. Calculate the partial pressure of hydrogen.

Answer - 19 : - Let the mass of H₂ in the mixture = 20 g
The mass of O₂ inthe mixture will be = 80 g

Question - 20 : - What would be SI units of a quantity pV²T²/n ?

Answer - 20 : -

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