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Ratling [72]
3 years ago
12

What is the vapor pressure of a solution in which the mole fraction of the solute is 0.200 and the vapor pressure of the pure so

lvent is 100.0 torr? (Assume a single nonvolatile, nonelectrolyte solute).
a. 0 torr
b. 80.0 torr
c. 100.0 torr
d. 120.0 torr
e. 20.0 torr
Chemistry
1 answer:
Ulleksa [173]3 years ago
7 0

Answer: The vapor pressure of a solution in which the mole fraction of the solute is 0.200 is 80.0 torr

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

\frac{p^o-p_s}{p^o}=i\times x_2

where,

p^0= vapor pressure of pure solvent = 100.0 torr

p_s = vapor pressure of solution = ?

i = Van'T Hoff factor = 1 for nonvolatile, nonelectrolyte solute

x_2 = mole fraction of solute  = 0.200  

\frac{100.0-p_s}{100.0}=1\times 0.200

p_s=80.0torr

The vapor pressure of a solution in which the mole fraction of the solute is 0.200 is 80.0 torr

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2 years ago
When an object is lifted 10 feet off the ground, it gains a certain amount of potential energy. If the same object is lifted 30
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Explanation:

Potential energy is the energy possessed by an object by virtue of its position.

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m= mass of object

g = acceleration due to gravity

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2 years ago
5.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the
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Explanation:

To calculate the entalpy, we use the equation:

q=mc\Delta T

where,

q = heat absorbed by water = ?

m = mass of water = {\text {volume of water}}\times {\text {density of water}}=100.0ml\times 1.00g/ml=100.0g

c = heat capacity of water = 4.186 J/g°C

\Delta T= change in temperature = 6.70^0C

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Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

The heat absorbed by water will be equal to heat released by MgSO_4

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Molar mass  = 120 g/mol

Putting values in above equation, we get:

\text{Moles of }MgSO_4=\frac{5.11g}{120g/mol}=0.042mol

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Answer:

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