Discussion Forum

18 g of glucose (C₆H₁₂O₆) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution

Answer:

Explanation:

Key concept :

Vapour pressure of water (p^o) = 760 torr

Number of moles of glucose = $\frac{Mass (g)}{Molecular mass (g/mol)}$

$=\frac{18 g}{180 gmol^{-1}}=0.1 mol$

Molar mass of water= 18 g/mol

Mass of water (given) = 178.2g

Number of moles of water= $\frac{Mass of water}{Molar mass of water}$

$=\frac{178.2}{18 g/mol}=9.9 mol$

Total number of moles = (0.1+ 9.9)moles = 10 moles

Now, mole fraction of glucose in solution = Change in pressure with respect to initial pressure

ie $\frac{\Delta p}{p^o}=\frac{0.1}{10}$

or $\Delta p = 0.01p^o$

$=0.01\times 760=7.6 torr$

Vapour pressure of solution = (760-7.6) torr

                                         = 752.4 torr