The vapour pressure of 30 % (w / v) , aqueous solution of glucose...

To calculate the vapour pressure of the $30$ (w/v) aqueous solution of glucose, we can use Raoult's law, which states that the vapour pressure of a component in an ideal solution is directly proportional to its mole fraction in the solution.

Weight of solution $= 100 \times 1 . 2 = 120gm$

Weight of water $= 120 - 30 = 90gm$

Now using formula

$\dfrac{{P}_{0} - P}{P} = \dfrac{molesof glucose}{molesof water}$

$\dfrac{24 - P}{P} = \dfrac{\dfrac{30}{180}}{\dfrac{90}{18}} = \dfrac{3}{90}$

$24 \times 90 - P \times 90 = 3P$

$⇒ P = 23 . 22$