What is the freezing point of a 4.00 molality Calcium Chloride?

Let’s learn what is the freezing point of a 4.00 molality Calcium Chloride. The most accurate or helpful solution is served by Yahoo! Answers.

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Answer:

Using the formula: ΔT=i(kf)m, the freezing point would be -22.32 °C

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Answer:

To calculate molarity, divide the number of moles of solute by the number of liters of solution.

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What is the freezing point of an aqueous 2.65 m calcium chloride (CaCl2) solution?

The freezing point of pure water is 0.0ºC and Kf of pure water is -1.86ºC/m. Note: Round your answer to one decimal place.

Answer:

dT = i x Kf x m, where dT = temperature lowering, i = van't Hoff factor, Kf = freezing point depression...

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1. What is the molality of ammonium ions in a 1.75 m solution of (NH4)3PO4? 2. What is the total molality of p?

1. What is the molality of ammonium ions in a 1.75 m solution of (NH4)3PO4? 2. What is the total molality of particles in the solution in question 1? 3. What is the total molality of particles in a 1.75 m solution of sugar (a covalent compound with the...

Answer:

1. 5.25m 2. 7.0 m 3. 1.75 m 4. -6.95 Degrees Celcius 5.a.) 108.4 Degrees Celcius b.) -30.4 Degrees Celcius...

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Which solution has highest freezing point if all are at the same molality?

which solution has highest freezing point if all are at the same molality calcium chloride sodium nitrate potassium sulfate iron(III) nitrate all have same freezing point please explain how you figure this out

Answer:

Sodium nitrate has the highest freezing point because it has the least number of ions (two) and therefore...

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Chemistry question about molality and stuff...plz help!!!? When freezing point depression and boiling point el?

Chemistry question about molality and stuff...plz help!!!? When freezing point depression and boiling point elevation are discussed, in most cases, the solute described is a non-ionic solute. Suggest an explanation for the following observation. A solution...

Answer:

depression in freezing point = i X Kf X m vant hoff factor , i = experimental value of colligative popery...

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Chemistry question on how to find molality, freezing point and boiling point.?

275g Sodium chloride is dissolved in 1.50 kg water a)find molality b)find new freezing point c)find new boiling point

Answer:

*** a *** 275 g NaCl x (1 mole / 58.45g) = 4.70 moles NaCl molality = moles solute / kg solvent = 4...

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