The vapour pressure of water at $23^{o} C$ is 19.8 mm. 0.1 mole glucose is dissolved in 178.2 g water. What is the vapour pressure (in mm) of the resultant solution?

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19.602

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19.402

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19.202

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Solution

The correct option is B 19.602
$n_{B} = 0.1, n_{A} = \frac{178.2}{18} = 9.9$
$x_{A} = \frac{n_{A}}{n_{A} + n_{B}} = \frac{9.9}{9.9 + 0.1} = 0.99$
$p = p_{0} x_{A}$
$= 19.8 \times 0.99 = 19.602 m m$

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The vapour pressure of water at 23oC is 19.8 mm. 0.1 mole glucose is dissolved in 178.2 g water. What is the vapour pressure (in mm) of the resultant solution?

The correct option is B 19.602nB=0.1,nA=178.218=9.9xA=nAnA+nB=9.99.9+0.1=0.99p=p0xA=19.8×0.99=19.602mm

The vapour pressure of water at $23^{o} C$ is 19.8 mm. 0.1 mole glucose is dissolved in 178.2 g water. What is the vapour pressure (in mm) of the resultant solution?

The correct option is B 19.602
$n_{B} = 0.1, n_{A} = \frac{178.2}{18} = 9.9$
$x_{A} = \frac{n_{A}}{n_{A} + n_{B}} = \frac{9.9}{9.9 + 0.1} = 0.99$
$p = p_{0} x_{A}$
$= 19.8 \times 0.99 = 19.602 m m$