# Find vapour pressure of 1 m aquaous solution of glucose at it's normal boiling point

$\frac{{P}^{0}-P}{{P}^{0}}=x\phantom{\rule{0ex}{0ex}}\u2206P=x.P\phantom{\rule{0ex}{0ex}}=\frac{1}{1+{\displaystyle \frac{1000}{18}}}\times 760\left(STP\right)\phantom{\rule{0ex}{0ex}}=13.44$

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