Two liquids were tested as described in Part 1 of the experiment. During evapora

Question

Two liquids were tested as described in Part 1 of the experiment. During evaporation. Liquid A underwent a change in temperature of -1.3 degree C and Liquid B underwent a temperature change of -2.1 degree C. Which liquid had the stronger intermolecular bonding?_____ Explain your reasoning: Calculate the mass of solid NaCl needed to add to 50.0 g of water to make 3.and 6.0m aqueous solutions. Recall that molality = moles solute 1 1000 g solvent, snow you calculations (include units). 3.0 m NaCl: _____ 4.5 m NaCl: _____ 6.0 m NaCl: _____

Explanation / Answer

3.

part-1: Given molality = 3.0 m, mass of the solvent = 50.0 g

1000 g of water contains NaCl = 3 moles

50 g of water contains NaCl = (3/1000) X 50 = 0.15 mol

Molecular mass of NaCl = 58.44 g/mol

Hence, mass of 0.15 mole of NaCl = 0.15 mol X 58.44 g/mol = 8.766 g

part-2: Given molality = 4.5 m, mass of the solvent = 50.0 g

1000 g of water contains NaCl = 4.5 moles

50 g of water contains NaCl = (4.5/1000) X 50 = 0.225 mol

Molecular mass of NaCl = 58.44 g/mol

Hence, mass of 0.15 mole of NaCl = 0.225 mol X 58.44 g/mol = 13.149 g

part-1: Given molality = 6.0 m, mass of the solvent = 50.0 g

1000 g of water contains NaCl = 6 moles

50 g of water contains NaCl = (6/1000) X 50 = 0.3 mol

Molecular mass of NaCl = 58.44 g/mol

Hence, mass of 0.15 mole of NaCl = 0.3 mol X 58.44 g/mol = 17.532 g

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