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Standard XII

Chemistry

Raoult's Law

1 mole of hep...

Question

1 mole of heptane (V.P. $= 92 m m$ of Hg) was mixed with 4 moles of octane (V.P $= 31$ mm of Hg). The vapour pressure of resulting ideal solution is:

46.2 mm of Hg

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40.0 mm of Hg

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43.2 mm of Hg

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38.4 mm of Hg

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Solution

The correct option is D 43.2 mm of Hg
As we know,

$P_{m} = P_{1}^{o} x_{1} + P_{2}^{o} x_{2}$

$x_{1} = \frac{1}{5}$

$x_{2} = \frac{4}{5}$

$P_{m} = 92 \times \frac{1}{5} + 31 \times \frac{4}{5}$

$= 18.4 + 24.8 = 43.2 m m H g$

Hence, the correct option is $C$

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1 mole of heptane (V.P. =92mm of Hg) was mixed with 4 moles of octane (V.P=31 mm of Hg). The vapour pressure of resulting ideal solution is:

The correct option is D 43.2 mm of HgAs we know,Pm=Po1x1+Po2x2x1=15x2=45Pm=92×15+31×45 =18.4+24.8=43.2 mm HgHence, the correct option is C

1 mole of heptane (V.P. $= 92 m m$ of Hg) was mixed with 4 moles of octane (V.P $= 31$ mm of Hg). The vapour pressure of resulting ideal solution is:

The correct option is D 43.2 mm of Hg
As we know,

$P_{m} = P_{1}^{o} x_{1} + P_{2}^{o} x_{2}$

$x_{1} = \frac{1}{5}$

$x_{2} = \frac{4}{5}$

$P_{m} = 92 \times \frac{1}{5} + 31 \times \frac{4}{5}$

$= 18.4 + 24.8 = 43.2 m m H g$

Hence, the correct option is $C$