A laboratory solution is prepared by completely dissolving of pure glucose () into of pure water. Wh — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

At $100^\circ\text{C}$ (the normal boiling point of water), the vapour pressure of pure water $P^0$ is $1\text{ atm} = 760\text{ Torr}$. Moles of glucose $n = \frac{18}{180} = 0.1\text{ mol}$. Moles of water $N = \frac{178.2}{18} = 9.9\text{ mol}$. Total moles = $9.9 + 0.1 = 10\text{ mol}$. The mole fraction of water $\chi_{water} = \frac{9.9}{10} = 0.99$. By Raoult's Law, $P_{solution} = P^0 \times \chi_{water} = 760 \times 0.99 = 752.4\text{ Torr}$.