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16 gm of oxygen and 3 gm of hydrogen are mixed and kept in 760 mm pressure at 0ºC. The total volume occupied by the mixture will be nearly
- a)22.4 l
- b)33.6 l
- c)448 litres
- d)44800 ml
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
16 gm of oxygen and 3 gm of hydrogen are mixed and kept in 760 mm pres...
°C. Find the total pressure of the mixture and the partial pressure of each gas.
First, we need to calculate the moles of each gas present in the mixture.
Moles of oxygen = 16 g / 32 g/mol = 0.5 mol
Moles of hydrogen = 3 g / 2 g/mol = 1.5 mol
The total moles of gas in the mixture = 0.5 + 1.5 = 2 mol
Now, we can use the ideal gas law to calculate the total pressure of the mixture.
PV = nRT
Where P = pressure, V = volume, n = moles, R = gas constant, and T = temperature.
We are given that the pressure is 760 mm and the temperature is 0°C, which is 273 K. We also assume that the volume is constant.
R = 0.0821 L·atm/mol·K
Ptotal = nRT / V
Ptotal = (2 mol) x (0.0821 L·atm/mol·K) x (273 K) / V
Ptotal = 44.8 / V
To find the value of V, we need to use the combined gas law.
P1V1 / T1 = P2V2 / T2
Since the volume is constant, we can simplify this to:
P1 / T1 = P2 / T2
We can use this equation to find the partial pressure of each gas.
For oxygen:
P1 = 760 mm
T1 = 273 K
T2 = 273 K
P2 = ?
P2 = P1 x (T2 / T1)
P2 = 760 mm x (273 K / 273 K)
P2 = 760 mm
For hydrogen:
P1 = 760 mm
T1 = 273 K
T2 = 273 K
P2 = ?
P2 = P1 x (T2 / T1)
P2 = 760 mm x (273 K / 273 K)
P2 = 760 mm
Therefore, the total pressure of the mixture is 44.8 / V mm, and the partial pressure of oxygen and hydrogen are both 760 mm.
First, we need to calculate the moles of each gas present in the mixture.
Moles of oxygen = 16 g / 32 g/mol = 0.5 mol
Moles of hydrogen = 3 g / 2 g/mol = 1.5 mol
The total moles of gas in the mixture = 0.5 + 1.5 = 2 mol
Now, we can use the ideal gas law to calculate the total pressure of the mixture.
PV = nRT
Where P = pressure, V = volume, n = moles, R = gas constant, and T = temperature.
We are given that the pressure is 760 mm and the temperature is 0°C, which is 273 K. We also assume that the volume is constant.
R = 0.0821 L·atm/mol·K
Ptotal = nRT / V
Ptotal = (2 mol) x (0.0821 L·atm/mol·K) x (273 K) / V
Ptotal = 44.8 / V
To find the value of V, we need to use the combined gas law.
P1V1 / T1 = P2V2 / T2
Since the volume is constant, we can simplify this to:
P1 / T1 = P2 / T2
We can use this equation to find the partial pressure of each gas.
For oxygen:
P1 = 760 mm
T1 = 273 K
T2 = 273 K
P2 = ?
P2 = P1 x (T2 / T1)
P2 = 760 mm x (273 K / 273 K)
P2 = 760 mm
For hydrogen:
P1 = 760 mm
T1 = 273 K
T2 = 273 K
P2 = ?
P2 = P1 x (T2 / T1)
P2 = 760 mm x (273 K / 273 K)
P2 = 760 mm
Therefore, the total pressure of the mixture is 44.8 / V mm, and the partial pressure of oxygen and hydrogen are both 760 mm.
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Community Answer
16 gm of oxygen and 3 gm of hydrogen are mixed and kept in 760 mm pres...
Pv=nRT
p=760mmHg=1atm
n=16/32+3/2=2
R=0.821
V=2×0.0821×273.15=44.8l=44800ml
p=760mmHg=1atm
n=16/32+3/2=2
R=0.821
V=2×0.0821×273.15=44.8l=44800ml
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16 gm of oxygen and 3 gm of hydrogen are mixed and kept in 760 mm pressure at 0ºC. The total volume occupied by the mixture will be nearlya)22.4 lb)33.6 lc)448 litresd)44800 mlCorrect answer is option 'D'. Can you explain this answer?
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