Solutions and Colligative PropertiesmediumMCQ SINGLE

Molal depression constant for Pb is . How many g of Sn must be dissolved in 425 g of Pb to produce aSolutions and Colligative Properties Chemistry Question

Question

Molal depression constant for Pb is $8.5^\circ\text{C kg/mol}$. How many g of Sn must be dissolved in 425 g of Pb to produce an alloy with melting point of $322^\circ\text{C}$? The melting point of pure Pb is $327^\circ\text{C}$. (Sn = 120)

Answer: A

💡 Solution & Explanation

The freezing point depression $\Delta T_f = 327 - 322 = 5^\circ\text{C}$. Using $\Delta T_f = K_f \times (w / (M \times W_{\text{kg}}))$, we have $5 = 8.5 \times (w / (120 \times 0.425))$. $5 = 8.5 \times w / 51 \Rightarrow 5 = w / 6 \Rightarrow w = 30.0 \text{ g}$. Therefore, correct answer is A.

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