Question

the vapour pressure of 10% aqueous glucose solution at 100 degrees C is

Answer 1

Answer:

Explanation:

Hope this helps you

Answer 2

Answer:

The vapour pressure is $752.4mm$

Explanation:

We know that $\frac{P_{0} -P_{S} }{P_{0} }$ $= x_{solute}$   --- Let this be (i)

Where $P_{0}$ is the vapour pressure of the solvent and $P_{S}$ is the vapour pressure of the solution.

At $100^{0} C$, vapour pressure of $H_{2} O = 760$

$x_{solute} = \frac{n_{solute} }{n_{solute+n_{solvent} } }$

Molecular weight of glucose $[C_{6} H_{12} O_{6} ]$ $=180$

Weight of glucose $=10gm$

Weight of $H_{2} O$ (water) $= 90gm$

By substituting the formula (i) : $\frac{P_{0} -P_{S} }{P_{0} }$ ,

the ratio of vapour pressure lowering of solvent from solution to the vapour pressure of pure solvent.

$\frac{760-P_{S} }{760} = \frac{\frac{10}{180} }{\frac{10}{180}+\frac{90}{18}} \\\\= \frac{0.05}{5.05} \\\\= 0.009$

From this, we can find the value of $P_{S}$ to be $752.4mm$.