The vapour pressure of pure water at is . If of pure glucose () is dissolved in of water, what will — Solutions and Colligative Properties Chemistry Question
Question
The vapour pressure of pure water at $20^\circ\text{C}$ is $17.5\text{ mm Hg}$. If $18\text{ g}$ of pure glucose ($C_6H_{12}O_6$) is dissolved in $178.2\text{ g}$ of water, what will be the equilibrium vapour pressure of the resulting solution?
Answer: A
💡 Solution & Explanation
Moles of glucose $= 18/180 = 0.1\text{ mol}$. Moles of water $= 178.2/18 = 9.9\text{ mol}$. Total moles $= 10$. Mole fraction of water $X_A = 9.9/10 = 0.99$. By Raoult's law, $P_s = P^0_A X_A = 17.5 \times 0.99 = 17.325\text{ mm Hg}$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes