A 2.100x102 M solution of NaCl in water is at 20.0° C. The sample was created by

Question

A 2.100x102 M solution of NaCl in water is at 20.0° C. The sample was created by dissolvinga sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 94 mL. The density of water at 20.0° C is 0.9982 g/mL. Part A Calculate the molality of the salt solution. Express your answer to four significant figuros and include the appropriate units. View Available Hint(s) mNaci = 1 Value Units Submit Part B Calculate the mole fraction of salt in this solution. Express the mole fraction to four significant figures > View Available Hint(s) cBookAi

Explanation / Answer

mass of the solution= density * volume = 0.9982 * 999.4 gm/mL .mL = 997.60 gm.= 0.997 kg

Molarity is 2.1 *10^-2 i.e. 2.1 *10^-2 moles of NaCl in one litre of solution. THus the weight of NaCl is 2.1 * 58.4 *10^-2 = 1.22 gm.

mass of solvent = 997.6-1.22 gm = 996.38 gm.= 0.9963 kg

Molality = moles of solute / mass of solvent = 2.1*10^-2 /0.9963 = 0.021 m.

Part B = Mole fraction of NaCl

= moles of NaCl / moles of water.

molality is moles of NaCl in 1 kg of water.

thus 0.021/1000 = 2.1 *10^-5 moles in 1 gm.

hence the mole fraction = 2.1 *10^-5 moles/gm * 58.4 gm/mole = 1.22*10^-3.

Part C

Molality = (mol of solute) / (kg of solution) = 0.021 molal.
% Mass = (Mass of solute) / (Mass of Solute + Mass of Solution) = 1.22 / (1.22 + 997.6) = 0.122

Part D

ppm = gm of solute /gm of solvent * 10^6 = 1.22/996.38 * 10^6 = 1220 ppm.

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