Solutions and Colligative PropertieshardMCQ SINGLE⭐ Must-Do

The relative decrease in vapour pressure of an aqueous solution is observed to be (approx. ). How maSolutions 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: $6(2n) = 2n + 10 \implies 12n = 2n + 10 \implies 10n = 10 \implies n = 1\text{ mole}$.

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