A salt solution was prepared by adding 410 g of NaCl to 4500 g of water at 32 de

Question

A salt solution was prepared by adding 410 g of NaCl to 4500 g of water at 32 degree C. MW of NaCl=58.44 g/mol. Determine: The mass fraction of sodium chloride in the solution. (0.0835) The molarity of sodium chloride in the solution. (1.506mol/L) The mole fraction of water in the solution. (0.973) The number of parts per million (ppm) of NaCl in the solution. (83503ppm) Use the following data to estimate the density of the solution. Density of NaCl solution in given in the following table in g/cm^3 or in SG

Explanation / Answer

1) Mass fraction = (mass of NaCl/mass of solution) = 410/4910 = 0.0835

2) volume of the solution = mass/density = 4910/1 = 4910 ml = 4.91 litres

moles of NaCl = mass/molar mass = 410/58.44 = 7.016

Thus, molarity = moles of NaCl/volume of solution in litres = 7.016/4.91 = 1.429 M

3) Moles of water = mass of water/molar mass of water = 4910/18 = 272.778

moles fraction of NaCl = moles of NaCl/(moles of water+moles of NaCl) = 7.016/(7.016+272.778) = 0.0251

Thus, mole fraction of water = 1 - mole fraction of NaCl = 0.975

4) ppm = (mass of NaCl/mass of solution)*106 = 8.35*104

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