What is the relative lowering in vapour pressure of 10% aqueous solution of glucose :

1 / 10

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1 / 90

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Solution

The correct option is C $1 / 90$
10% aqueous solution of glucose

10 gm glucose in 100 gm solution

so $H_{2} O$ will be 90 gm

$n_{g l u c o s e} = \frac{10}{180} = \frac{1}{18}$

$n_{H_{2} O} = \frac{90}{18} = 5$

$X_{g l u c o s e} = \frac{\frac{1}{18}}{5 + \frac{1}{18}} = \frac{1}{90}$

so $\frac{Δ P}{P_{s o l v}^{O}} = X_{g l u c o s e}$

$= \frac{1}{90}$
Hence, the correct option is $B$

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What is the relative lowering in vapour pressure of 10% aqueous solution of glucose :

The correct option is C 1/9010% aqueous solution of glucose10 gm glucose in 100 gm solutionso H2O will be 90 gm nglucose=10180=118nH2O=9018=5Xglucose=1185+118=190so ΔPPOsolv=Xglucose=190Hence, the correct option is B