What is the freezing point for a solution containing 1.00 mole of NaCl (an ionic component) dissolved in 1.00 kg of water?
Group of answer choices
+3.72 oC
-3.72 oC
-1.86 oC
+1.86 oC
What is the freezing point for a solution containing 1.00 mole of NaCl (an ionic component)...
Which of the following solute amounts, when dissolved in 1.00 kg of water, would produce a solution with a freezing point of -1.86 oC? Group of answer choices 2.00 moles of sugar 0.50 moles of KBr 2.00 moles of NaI 1.00 mole of NaCl
An aqueous solution containing sodium chloride has a freezing point that is -1.28 degrees C at 1.0 atm. What is the % by mass of NaCL in this solution? HINT: The Kf for water = 1.86 degrees C kg/mol
What is the freezing point of an aqueous 1.00 m NaCl solution? (Kf = 1.86 *C/m) (Assume complete dissociation of the salt.) I know the answer is -3.78*C but I cannot figure out how to get it. I keep getting -1.86*C. Please show step by step so I can understand. Thank you.
You have a aqueous solution of NaCl that has a freezing point of -9.35°C. Assuming a van't Hoff factor of 1.9 for NaCl, what is the mass percent of chloride ion in the solution? (Kf for water is 1.86°C kg/mol).
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A solution containing 1.00 g of an unknown non-electrolyte liquid and 9.00 g water has a freezing point of -3.33 oC. The Kf = 1.86 oC/m for water. Calculate the molar mass of the unknown liquid, in g/mol.
The freezing point of a solution that contains 1.00 g of an unknown compound, (A), dissolved in 10.0 g of benzene is found to be 2.17 oC. The freezing point of pure benzene is 5.48 oC. The molal freezing point depression constant of benzene is 5.12 oC/molal. What is the molecular weight of the unknown compound?
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Determine the melting point of an aqueous solution containing 147 mg of saccharin (C7H5O3NS) added to 1.00 mL of water (density of water = 1.00 g/mL, Kf = 1.86°C/m). What is the freezing point of a 1.276 m aqueous solution of aluminum nitrate, Al(NO3)3? The Kfof water is 1.86°C/m.
Sodium chloride (NaCl) is commonly used to melt ice on roads during the winter. Calcium chloride (CaCl2) is sometimes used for this purpose too. Let us compare the effectiveness of equal masses of these two compounds in lowering the freezing point of water by calculating the freezing point depression of solutions containing 230. g of each salt in 1.00 kg of water. (An advantage of CaCl2 is that it acts more quickly because it is hygroscopic, that is, it absorbs...