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If 10-4 dm3...

Question

If $10^{- 4} {d m}^{3}$ of water is introduced into a $1.0 {d m}^{3}$ flask at $300 K$ , then how many moles of water are in the vapour phase when equilibrium is established?
(given vapour pressure of $H_{2} O$ at $300 K$ is $3170 P a; R = 8.314 J K^{- 1}$ ${m o l}^{- 1}$ )

5.56 \times 10^{- 3} m o l

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1.53 \times 10^{- 2} m o l

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4.46 \times 10^{- 2} m o l

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1.27 \times 10^{- 3} m o l

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Solution

The correct option is D $1.27 \times 10^{- 3} m o l$
The volume occupied by water molecules in vapour phase is $(1 \times 10^{- 4})$ $d m^{3}$ , that is approximately $(1 \times 10^{- 3}) m^{3}$
$p_{v a p} V = n_{H_{2} O}$ mol

$3170 \times 1 \times 10^{- 3} = n_{H_{2} O} \times 8.314 \times 300 K$

$n_{H_{2} O} = \frac{3170 \times 1 \times 10^{- 3}}{8.314 \times 300} = 1.27 \times 10^{- 3}$ mol

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If $10^{- 4} d m^{3}$ of water is introduced into a $1 d m^{3}$ flask at 300 K, how many moles of water are in the vapour phase when equilibrium is established. (Given vapour pressure of $H_{2} O$ at 300 K is 3170 Pa; R = 8.314 J/K/mol)

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If 10−4 dm3 of water is introduced into a 1.0 dm3 flask at 300 K, then how many moles of water are in the vapour phase when equilibrium is established?(given vapour pressure of H2O at 300 K is 3170Pa;R=8.314J K−1 mol−1)

The correct option is D 1.27×10−3 molThe volume occupied by water molecules in vapour phase is (1×10−4) dm3 , that is approximately (1×10−3)m3 pvapV=nH2Omol3170×1×10−3=nH2O×8.314×300KnH2O=3170×1×10−38.314×300=1.27×10−3mol

If $10^{- 4} {d m}^{3}$ of water is introduced into a $1.0 {d m}^{3}$ flask at $300 K$ , then how many moles of water are in the vapour phase when equilibrium is established?
(given vapour pressure of $H_{2} O$ at $300 K$ is $3170 P a; R = 8.314 J K^{- 1}$ ${m o l}^{- 1}$ )