How much would the freezing point of water decrease if 4 mol of NaCl were added to 1 kg of water (Kf = 1.86°C/(mol/kg) for water and i = 2 for NaCl)?

A. 14.88°C
B. 3.72°C
C. 7.44°C
D. 1.86°C

PLEASE ANSWER ASAP

Freezing point decrease is a colligative property.

Freezing point decrease is given by the formula:

ΔfT = Kf * i * m

m = # moles solute / Kg solvent

# moles solute = 4 mol

Kg solvent = 1 Kg

=> m = 4 mole / 1 Kg = 4m

=> ΔTf = 1.86 °C / m * 2 * 4 m = 14.88 °C

Answer: 14.88°C

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sTurn sTurn · Answer 2 · 2018-12-17T18:13:33+01:00

sTurn sTurn

17-12-2018

14.88°C the answer would be A.

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How much would the freezing point of water decrease if 4 mol of NaCl were added to 1 kg of water (Kf = 1.86°C/(mol/kg) for water and i = 2 for NaCl)? A. 14.88°C B. 3.72°C C. 7.44°C D. 1.86°CPLEASE ANSWER ASAP

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How much would the freezing point of water decrease if 4 mol of NaCl were added to 1 kg of water (Kf = 1.86°C/(mol/kg) for water and i = 2 for NaCl)?

A. 14.88°C
B. 3.72°C
C. 7.44°C
D. 1.86°C

PLEASE ANSWER ASAP