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Center of Mass as an Average Point

The vapour pr...

Question

The vapour pressure of a dilute aqueous solution of glucose is 750 mm of Hg at 373 K. Calculate molality.

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Solution

$P^{o} = 760; P_{s} = 750 m m$
$\frac{P^{o} - P_{s}}{P^{o}} = χ_{2}$
$\Rightarrow \frac{760 - 750}{760} = χ_{2} \Rightarrow χ_{2} = 0.013; χ_{1} = 1 - 0.013 = 0.968$
$m = \frac{χ_{2} \times 1000}{χ_{1} \times M w_{1}} = \frac{0.013 \times 1000}{0.986 \times 18} = 0.732$

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