The relative decrease in vapour pressure of an aqueous solution is observed to be (approx. ). How ma — Solutions and Colligative Properties Chemistry Question
Question
The relative decrease in vapour pressure of an aqueous $NaCl$ solution is observed to be $0.167$ (approx. $1/6$). How many exact moles of $NaCl$ are present in $180\text{ g}$ of the water solvent? (Assume $100\%$ ionization).
Answer: B
💡 Solution & Explanation
RLVP $= \frac{i \times n}{i \times n + N}$. $NaCl$ yields $i=2$. Moles of water $N = 180/18 = 10$. Thus, $0.167 \approx 1/6 = \frac{2n}{2n + 10}$. Cross multiplying: $6(2n) = 2n + 10 \implies 12n = 2n + 10 \implies 10n = 10 \implies n = 1\text{ mole}$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes