2.34    Vapour pressure of water at  293 \; K is  17.535\; mm\; Hg. Calculate the vapour pressure of water at 293 \; K  when  25 \; g  of glucose is dissolved in 450 \; g of water.

Answers (1)

Firstly we will find number of moles of both water and glucose.

Moles of glucose :

                                 = \frac{25}{180} = 0.139\ mol                                        (Molar\ mass\ of\ glucose = 6(12)+12(1)+6(16) = 180\ g\ mol^{-1})

and moles of water :

                                   = \frac{450}{18} = 25\ mol                                            (Molar\ mass\ of\ water = 18\ g\ mol^{-1})

Now,

                                             \frac{p_w^{\circ} - p}{p_w^{\circ}} = \frac{n_g}{n_g + n_w}

or                                        \frac{17.535 - p}{17.535} = \frac{0.139}{0.139+ 25}

or                                          p = 17.44\ mm\ of\ Hg 

Thus vapour pressure of water after glucose addition = 17.44 mm of Hg

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