JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
How much amount of \(\mathrm{NaCl}\) should be added to \(600\; \mathrm{g}\) of water \((\rho=1.00\; \mathrm{g} / \mathrm{mL})\) to decrease the freezing point of water to \(-0.2^{\circ} \mathrm{C} ?\) ............. \(\mathrm{gm}\) (The freezing point depression constant for water \(=2\; \mathrm{K}\; \mathrm{kg} \;\mathrm{mol}^{-1}\) )
- A \(2.25\)
- B \(2\)
- C \(1.75\)
- D \(1.5\)
Answer & Solution
Correct Answer
(C) \(1.75\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{m} \times \mathrm{K}_{\mathrm{f}}\) \(0.2=2 \times 2 \times \frac{\mathrm{w} / 58.5}{600 / 1000}\) \(\mathrm{w}=1.755 \mathrm{gm}\)
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