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Yanka [14]
3 years ago
12

An enclosed vessel contains 2.5g of 9b nitrogen and 13.3g of chlorine at s.T.P. Of What will be the partial pressure of the Il n

itrogen if the temperature is raised to b 180C
Chemistry
1 answer:
kow [346]3 years ago
8 0

Answer:

0.535 atm

Explanation:

Since the volume of the tank is constant, we use Gay- Lussac's law to find the pressure at 180°C.

So, P₁/T₁ = P₂/T₂ where P₁ = pressure at S.T.P = 1 atm, T₁ = temperature at S.T.P = 273.15 K, P₂ = pressure of gas at 180 °C and T₂ = 180 °C = 273.15 + 180 K = 453.15 K

So, P₁/T₁ = P₂/T₂

P₂ = P₁T₂/T₁

Substituting the values of the variables into the equation, we have

P₂ = P₁T₂/T₁

P₂ = 1 atm × 453.15 K/273.15 K

P₂ = 1 atm × 1.66

P₂ = 1.66 atm

We now need to find the total number of moles of each gas present

number of moles of nitrogen = mass of nitrogen, m/molar mass of nitrogen molecule M

n = m/M

m = 2.5 g and M = 2 × atomic mass of nitrogen (since it is diatomic) = 2 × 14 g/mol = 28 g/mol

So, n = 2.5 g/28 g/mol

n = 0.089 mol

number of moles of chlorine, n' = mass of chlorine, m'/molar mass of chlorine molecule M'

n' = m'/M'

m' = 13.3 g and M = 2 × atomic mass of chlorine (since it is diatomic) = 2 × 35.5 g/mol = 71 g/mol

So, n' = 13.3 g/71 g/mol

n' = 0.187 mol

So, the total number of moles of gas present is n" = n + n' = 0.089 mol + 0.187 mol = 0.276 mol

So, the partial pressure due to nitrogen gas, P = mole fraction of nitrogen × pressure of gas at 180 °C

P = n/n" × P₂

P = 0.089 mol/0.276 mol × 1.66 atm

P = 0.322 × 1.66 atm

P = 0.535 atm

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25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.

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CAVA/CBVB = NA/NB

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