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Question: If \({{10}^{-4}}d{{m}^{3}}\) of water is introduced into a \(1.0d{{m}^{3}}\) flask at 300K, then how...

If 10−4dm3{{10}^{-4}}d{{m}^{3}} of water is introduced into a 1.0dm31.0d{{m}^{3}} flask at 300K, then how many moles of water are in the vapour phase when equilibrium is established?
(given vapour pressure of H2O{{H}_{2}}O at 300 K is 3170 Pa; R=8.314JK−1mol−1R = 8.314J{{K}^{-1}}mo{{l}^{-1}})
A. 5.56×10−3mol5.56\times {{10}^{-3}}mol
B. 1.53×10−2mol1.53\times {{10}^{-2}}mol
C. 4.46×10−2mol4.46\times {{10}^{-2}}mol
D. 1.27×10−3mol1.27\times {{10}^{-3}}mol

Explanation

Solution

To solve this question we have to use the Ideal gas equation that is PV = nRT. In the question vapour pressure of water is given so, the relation between the vapour pressure and ideal gas of water is PvapV=nvapRT{{P}_{vap}}V = {{n}_{vap}}RT. Convert the values of volume in the S.I unit from the C.G.S unit. 1dm = 0.1 m

Complete Solution :
As you have learned about the Ideal gas law in your chemistry lessons, Ideal gas is also known as perfect gas. This law establishes a relationship between pressure, volume, temperature and amount of gas(n) and they are four variables of gases.
Ideal gas equation is written as,
PV = nRT
Where P = Pressure of the gas
V = Volume of the gas
n = moles or amount of gas
R = universal gas constant
T = temperature.
If we have to write this equation in term of vapour pressure of then we write it as,
PvapV=nH2ORT{{P}_{vap}}V = {{n}_{{{H}_{2}}O}}RT ……..(1)
Where, Pvap{{P}_{vap}} = vapor pressure of water
nH2O{{n}_{{{H}_{2}}O}} = moles of water

- In the question it is told that we have to find the moles of water in vapour phase, and the values that are given in the question are,
Pvap{{P}_{vap}} = 3170 Pa
V =10−4dm3=10−3m3{{10}^{-4}}d{{m}^{3}}={{10}^{-3}}{{m}^{3}}
T = 300 K
R=8.314JK−1mol−1R = 8.314J{{K}^{-1}}mo{{l}^{-1}}
Now put all the values in formula 1, you will get,
3170×10−3=nH2O×8.314×3003170\times {{10}^{-3}}={{n}_{{{H}_{2}}O}}\times 8.314\times 300
⇒nH2O=3170×10−38.314×300=1.27×10−3mol\Rightarrow {{n}_{{{H}_{2}}O}}=\dfrac{3170\times {{10}^{^{-3}}}}{8.314\times 300}=1.27\times {{10}^{-3}}mol
So, the correct answer is “Option D”.

Note: Change the units wherever needed and change according to the requirement. Vapour pressure is known as pressure of a vapour when it comes in contact with its solid or liquid form. The value of universal gas constant also changes according to the units.