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RSB [31]
3 years ago
9

Cinnamon owes its flavor and odor to cinnamaldehyde (C9H8O). Determine the boiling point elevation of a solution of 92.7 mg of c

innamaldehyde dissolved in 1.00 g of carbon tetrachloride (Kb = 5.03°C/m).
Chemistry
1 answer:
Flauer [41]3 years ago
6 0

Answer:

The boiling point elevation is 3.53 °C

Explanation:

∆Tb = Kb × m

∆Tb is the boiling point elevation of the solution

Kb is the molal boiling point elevation constant of CCl4 = 5.03 °C/m

m is the molality of the solution is given by moles of solute (C9H8O) divided by mass of solvent (CCl4) in kilogram

Moles of solute = mass/MW =

mass = 92.7 mg = 92.7/1000 = 0.0927 g

MW = 132 g/mol

Moles of solute = 0.0927/132 = 7.02×10^-4 mol

Mass of solvent = 1 g = 1/1000 = 0.001 kg

m = 7.02×10^-4 mol ÷ 0.001 kg = 0.702 mol/kg

∆Tb = 5.03 × 0.702 = 3.53 °C (to 2 decimal places)

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If a gas occupies 4600 mL at 0.9 atm and 195°C, what is the new volume in ml
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Answer:

The new volume is 2415 mL

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases.

Boyle's law says that the volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure and is expressed mathematically as:

P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

\frac{V}{T} =k

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. This can be expressed mathematically in the following way:

\frac{P}{T} =k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

\frac{P*V}{T} =k

Having two different states, an initial state and an final state, it is true:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

In this case:

  • P1= 0.9 atm
  • V1=4,600 mL= 4.6 L (being 1 L=1,000 mL)
  • T1= 195 °C= 468 °K (being 0°C=273°K)

The final state 2 is in STP conditions:

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  • V2= ?
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Replacing:

\frac{0.9 atm*4.6L}{468K} =\frac{1 atm*V2}{273K}

Solving:

V2=\frac{0.9 atm*4.6L}{468K}*\frac{273K}{1 atm}

V2= 2.415 L =2,415 mL

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