45.8 g of NaCl is dissolved in 500. mL of a water (d = 1.00 g/mL) and is placed

Question

45.8 g of NaCl is dissolved in 500. mL of a water (d = 1.00 g/mL) and is placed in the freezer at -1 °C overnight. Which statement best describes what would happen to the solution?

The freezing point of the NaCl solution would be -5.83 °C, so the solution would not freeze.

The freezing point of the NaCl solution would be 5.83 °C, so the solution would freeze.

The freezing point of the NaCl solution would be -2.92 °C, so the solutio

The freezing point of the NaCl solution would be -5.83 °C, so the solution would not freeze.

The freezing point of the NaCl solution would be 5.83 °C, so the solution would freeze.

The freezing point of the NaCl solution would be -2.92 °C, so the solutio

Explanation / Answer

adding NaCl will decrease the Freezing point, so ignore B

dTf = -Kf*m*i

i = 2 ions

Kf = 1.86

m = mol/kg

mol NACl = mass/MW = 45.8/58 = 0.7896

kg = 500 g = 0.5 kg

then molality = 0.7896/0.5 = 1.5792

dTf = -Kf*m*i

dTf = -Kf*m*i = -1.86*1.5792*2 = -5.8746

choose A

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