Homework Answers
The depression in freezing point = Kf * molality, where Kf is freezing poiny constant of water with value 1.86 0C / m
0.3 = 1.86 * m
Solving, molality = 0.161 m
Moles of glucose / mass of water in kg = molality
moles of glucose = 0.161 * 0.228 = 3.67 * 10-2 moles
Mass = moles * molar mass = 3.67 * 10-2 * 180.2 = 6.613 g
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The melting point of water H2O is 0.00 °C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is antifreeze (ethylene glycol) . How many grams of antifreeze , CH2OHCH2OH ( 62.10 g/mol), must be dissolved in 289.0 grams of water to reduce the melting point by 0.350 °C ?
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References Use the References to access important values if needed for this question. The normal boiling point of chloroform (CHCI3) is 61.70 °C and its Kbp value is 3.67 °C/m. A nonvolatile, nonelectrolyte that dissolves in chloroform is aspirin. If 12.80 grams of aspirin, C,H,04 (180.1 g/mol), are dissolved in 284.5 grams of chloroform: The molality of the solution is The boiling point of the solution is °C Submit Answer
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References Use the References to access important values if needed for this question. The normal freezing point of water, H,0 is 0.00 °C and its Kfp value is 1.86 °C/m. Assuming complete dissociation of the electrolyte, if 14.42 grams of are dissolved in 183.4 grams of water what is the freezing point of the solution? ammonium bromide (NH, Br, 97.95 g/mol) oc Submit Answer
The melting 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,1,0, (227.1 g/mol), must be dissolved in 245.0 grams of benzene to reduce the melting point by 0.350 "C? g TNT
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