8 of 8 A 2.800x10-2 M solution of NaCI in water is at 20.0°C. The sample was cre

Question

8 of 8 A 2.800x10-2 M solution of NaCI in water is at 20.0°C. The sample was created by dissolving a 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 999.2 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 figures and include the appropriate units. View Available Hint s mNaci = 1.028 1111 Submit Incorrect. Try Again Part B

Explanation / Answer

number of moles of NaCl = 2.8*10^-2 * 1= 2.8 * 10^-2 moles

mass of NaCl = 2.8*10^-2 * 58.5 = 1.638 g

Volume of water required = 999.2 mL

density of water = 0.9982 g/mL

1mL of water has 0.9982 g

999.2 mL of water has 0.9982*999.2 = 997.4014 g

molality of the solution = mass of the solute/molar mass of solute * 1000/mass of the solvent

m = 1.638/58.5*1000/997.4104

m = molality = 0.0281 m

mole fraction of salt = number of moles of salt/total number of moles of solution

mole fraction = (1.638/58.5)/((1.638/58.5)+(997.4014/18)) = 5.05 * 10^-4

mass percentage of NaCl solution = mass of NaCl/mass of H2O * 100

% mass = 1.638/997.4014*100 = 0.16%

From calculated values mass of NaCl = 1.638g present in 1 L of solution

1 ppm = 1 mg/L

concentration of NaCl in ppm = 1.638g/L = 1638 mg/L

NaCL concentration = 1638 ppm

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