The freezing point of benzene C,H, is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is TNT (trinitrotoluene). How many grams of TNT, C,H&N3O6 (227.1 g/mol), must be dissolved in 269.0 grams of benzene to reduce the freezing point by 0.400°C ? Refer to the table for the necessary boiling or freezing point constant. Solvent Formula K CC/m) KrCC/m) Water H,0 0.512 1.86 Ethanol CH;CH,OH 1.22 1.99 Chloroform CHCI3 3.67 Benzene CH6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02
The freezing point of benzene C,H, is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is TNT (trinitrotoluene). How many grams of TNT, C,H&N3O6 (227.1 g/mol), must be dissolved in 269.0 grams of benzene to reduce the freezing point by 0.400°C ? Refer to the table for the necessary boiling or freezing point constant. Solvent Formula K CC/m) KrCC/m) Water H,0 0.512 1.86 Ethanol CH;CH,OH 1.22 1.99 Chloroform CHCI3 3.67 Benzene CH6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02
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The freezing point of benzene C,H, is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is TNT (trinitrotoluene).
How many grams of TNT, C,H<N30, (227.1 g/mol), must be dissolved in 269.0 grams of benzene to reduce the freezing point by 0.400°C ? Refer to
the table for the necessary boiling or freezing point constant.
Solvent
Formula
K, (°C/m) Kr (°C/m)
Water
H2O
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,CH3
2.02
g TNT.
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Expert Solution
Step 1
Given:
Depression in freezing point =
Molar mass of TNT = 227.1 g/mol
Mass of benzene = 269.0 g
Kf of benzene =
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Solved in 3 steps
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