The freezing point of water is 0.00°C at 1 atmosphere. A student dissolves 12.27 grams of sodium acetate, NaCH3C00 (82.00 g/mol), in 174.4 grams of water. Use the table of boiling and freezing point constants to answer the questions below. Solvent Formula K, °C/m) KfCC/m) Water H20 0.512 1.86 Ethanol CH;CH,OH 1.22 1.99 Chloroform CHCI3 3.67 Benzene CH6 2.53 5.12 Diethyl ether CH3CH;0CH,CH; 2.02 The molality of the solution is | m. The freezing point of the solution is °C.

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The freezing point of water is 0.00°C at 1 atmosphere.
A student dissolves 12.27 grams of sodium acetate, NaCH3co0 (82.00 g/mol), in 174.4 grams of water. Use the table of
boiling and freezing point constants to answer the questions below.
Formula
K, "C/m) Kf ("C/m)
Solvent
Water
H20
0.512
1.86
Ethanol
CH;CH2OH
1.22
1.99
Chloroform
CHC13
3.67
Benzene
2.53
5.12
Diethyl ether CH3CH2OCH2CH3
2.02
The molality of the solution is
m.
The freezing point of the solution is
°C.
Transcribed Image Text:The freezing point of water is 0.00°C at 1 atmosphere. A student dissolves 12.27 grams of sodium acetate, NaCH3co0 (82.00 g/mol), in 174.4 grams of water. Use the table of boiling and freezing point constants to answer the questions below. Formula K, "C/m) Kf ("C/m) Solvent Water H20 0.512 1.86 Ethanol CH;CH2OH 1.22 1.99 Chloroform CHC13 3.67 Benzene 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02 The molality of the solution is m. The freezing point of the solution is °C.
Expert Solution
Step 1

Given 

Mass of Sodium acetate =  12.27 gram 

Molar mass of sodium acetate =  82 gram/mole

Mass of water = 174.4 gram

                       = 0.1744 Kg                   ( 1 gram = 10-3 Kg ) 

Freezing point of water = 0 ℃ 

Molality of solution = ? 

Freezing point of solution = ?

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