Calculate the freezing point (in degrees C) of asolution made by dissolving 4.08

Question

Calculate the freezing point (in degrees C) of asolution made by dissolving 4.08 g of naphthalene{C10H8} in 94.0 g of CCl4. TheKfp of the solvent is 29.8 K/m and the normal freezingpoint is -23 degrees C.

Correct Answer: -33.11

Please Help me! thanks! Calculate the freezing point (in degrees C) of asolution made by dissolving 4.08 g of naphthalene{C10H8} in 94.0 g of CCl4. TheKfp of the solvent is 29.8 K/m and the normal freezingpoint is -23 degrees C.

Correct Answer: -33.11

Please Help me! thanks!

Explanation / Answer

molar mass = 10*12 + 8*1 = 128 g/mol molality = (4.08 g Naph * 1mol / 128 g) / (0.094 kg CCl4) =0.339095745 m T = Kf*m = (29.8 K/m) *(0.339 m) = 10.1050532 C Tf, new = Ti - T = -23C - 10.105 C = -33.1050532 C Tf, new ~ -33.11 C

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