3) Salt, NaCI, has a solubility of 35.7 g per 100 g of water at 0°C. The solubil

Question

3) Salt, NaCI, has a solubility of 35.7 g per 100 g of water at 0°C. The solubility of salt in water does not vary appreciably with temperature. NaCl is an electrolyte that commonly used to melt ice by lowering its melting point. What is the lowest possible melting point for ice that could be obtained considering the solubility of NaCI in water? The K, for H,O is 1.86°C/m. 4. A 1.2436 g sample of an unknown nonelectrolyte is dissolved in 10.9303 g of benzophenone produces a solution that freezes at 43.2°C. If the pure benzophenone melted at 48.1°C, what is the molecular weight of the unknown compound? The Kfor benzophenone is 9.8°C/m.

Explanation / Answer

1) The solubility of a substance expresses how much of it dissolves in a certain amount of solvent.

For NaCl: 35.7 gr NaCl / 100 gr H2O Since it is a maximum dilution value, we determine the molality of said substance with those same characteristics: 35.7 gr NaCl / 58.44 gr / mol = 0.611 mol NaCl 100 gr H2O * (1Kg / 1000 g) = 0.1 Kg m = mol / Kg = 0.611 / 0.1 = 6.11 m Determining the maximum temperature delta: ?T = K * m = 1.86 * 6.11 = 11.36 ºC Tc = 0ºC - 11.36ºC = -11.36ºC 2) Calculating the ?T that originated the solution we have: ?T = 48.1 - 43.2 = 4.9 ºC of the ?T equation we clear the required molality: m = ?T / K = 4.9 / 9.8 = 0.5 mol / Kg We find the moles represented by the aggregate mass: 0.5 mol sto / Kg ste * (10.9303 g / 1000 Kg / g) = 5.46515 * 10 ^ -3 mol sto MW = solute mass / solute mol = 1.2436 gr / 5.46515 * 10 ^ -3 mol = 227.51 gr / mol.

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