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otez555 [7]
3 years ago
5

when salt is added to water how does it affect the water's freezing points? Explain the science behind how this works at the par

ticle level for water and salt?
Chemistry
1 answer:
Sphinxa [80]3 years ago
3 0
Under normal conditions, water freezes at 0 degrees Celsius or 32 degrees Fahrenheit. However by, adding salt, the water will cause the freezing point to become lower.
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Which substance is not a structural isomer of hexyne?
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c) 3,3-dimethylpent-1-yne

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2 years ago
When 231. mg of a certain molecular compound X are dissolved in 65. g of benzene (CH), the freezing point of the solution is mea
Elan Coil [88]

Answer:

Molar mass X = 18.2 g/mol

Explanation:

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Mass of compound X = 231 mg = 0.231 grams

Mass of benzene = 65.0 grams

The freezing point of the solution is measured to be 4.5 °C.

Freezing point of pure benzene = 5.5 °C

The freezing point constant for benzene is 5.12 °C/m

Step 2: Calculate molality

ΔT = i*Kf*m

⇒ΔT = the freezing point depression = 5.5 °C - 4.5 °C = 1.0 °C

⇒i = the Van't hoff factor = 1

⇒Kf = the freezing point depression constant for benzene = 5.12 °C/m

⇒m = the molality = moles X / mass benzene

m = 1.0 / 5.12 °C/m

m = 0.1953 molal

Step 3: Calculate moles X

Moles X = molality * mass benzene

Moles X = 0.1953 molal * 0.065 kg

Moles X = 0.0127 moles

Step 4: Calculate molar mass X

Molar mass X = mass / moles

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3 years ago
how many moles of iodine should be added to 750 grams of carbon tetrachloride to prepare a 0.24 m solution
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Answer:

0.18 mol

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Given data

  • Mass of carbon tetrachloride (solvent): 750 g
  • Molality of the solution: 0.24 m
  • Moles of iodine (solute): ?

Step 1: Convert the mass of the solvent to kilograms

We will use the relationship 1 kg = 1,000 g.

750g \times \frac{1kg}{1,000g} =0.750kg

Step 2: Calculate the moles of the solute

The molality is equal to the moles of solute divided by the kilograms of solvent. Then,

m = \frac{moles\ of\ solute }{kilograms\ of\ solvent} \\moles\ of\ solute = m \times kilograms\ of\ solvent = \frac{0.24mol}{kg}  \times 0.750kg = 0.18 mol

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