Calculate the weight of a substance 'A' which in the gaseous form occupies 10 litres at 27°C and 700 mm pressure. The molecular weight of 'A' is 60. (Leave your answer as an unsimplified fraction.)
Ans.=22.45 g
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Answered by
2
Use Pv=nRT
Put n=w/m
Here w=weight of substance
M=molecular mass
You have to calculate w.
Remember changing Celsius to kelvin
Hope you get your answer.
Put n=w/m
Here w=weight of substance
M=molecular mass
You have to calculate w.
Remember changing Celsius to kelvin
Hope you get your answer.
ameyashrivastav:
not correct
its absolutely correct
Answered by
3
V = volume = 10 liters, T = 27°C = 300 °K.
P = 700 mm = 700/760 atm
R = 0.08205 atm-litre/mole/°K
We assume that the given gas is an ideal gas and so, we use the Ideal Gas law.
P V = n R T
700/760 * 10 = n * 0.08205 * 300
n = 0.3741 moles
Molecular weight of A = W = 60
Mass of the given substance A = n * W
= 0.3741 * 60 = 22.446 gms
Universal gas constant R can be expressed in different units too. We can convert all quantities like P & V into SI system and then use R = 8.314 J/K/mole.
P = 700 mm/760 mm * 1 atm = 700/760 * 1.013 * 10⁵ Pa
V = 10 liters = 10 * 10⁻³ m³
P = 700 mm = 700/760 atm
R = 0.08205 atm-litre/mole/°K
We assume that the given gas is an ideal gas and so, we use the Ideal Gas law.
P V = n R T
700/760 * 10 = n * 0.08205 * 300
n = 0.3741 moles
Molecular weight of A = W = 60
Mass of the given substance A = n * W
= 0.3741 * 60 = 22.446 gms
Universal gas constant R can be expressed in different units too. We can convert all quantities like P & V into SI system and then use R = 8.314 J/K/mole.
P = 700 mm/760 mm * 1 atm = 700/760 * 1.013 * 10⁵ Pa
V = 10 liters = 10 * 10⁻³ m³
:-)
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