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The complete question is as follows: 171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in one liter of water at 25 oC. At 25 oC the vapor pressure of water is 24 mmHg. Which value is closest to the vapor pressure (VP) of this solution at 25 oC?
a. 16mm Hg
b. 24mm Hg
c. 20mm Hg
d. 12mm Hg
Answer: The vapor pressure (VP) of this solution at is closest to the value 24 mm Hg.
Explanation:
Given: Mass of sucrose = 171 g
Mass of water = 1 L = 1000 g
Vapor pressure of water = 24 mm Hg
As moles is the mass of substance divided by its molar mass. Hence, moles of water (molar mass = 18.02 g) is calculated as follows.
Similarly, moles of sucrose (molar mass = 342 g/mol) is as follows.
Total moles = 55.49 + 0.5 mol = 55.99 mol
Mole fraction of water is as follows.
Formula used to calculate vapor pressure of the solution is as follows.
where,
= vapor pressure of component i over the solution
= vapor pressure of pure component i
= mole fraction of i
Substitute the values into above formula to calculate vapor pressure of water as follows.
Thus, we can conclude that the vapor pressure (VP) of this solution at is closest to the value 24 mm Hg.
Answer:
The amount of heat to absorb is 6,261 J
Explanation:
Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.
The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.
So:
- Heat required to raise the temperature of ice from -20 °C to 0 °C
Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).
In this case, m= 10 g, specific heat of the ice= 2.1 and ΔT=0 C - (-20 C)= 20 C
Replacing: Q= 10 g*2.1 *20 C and solving: Q=420 J
- Heat required to convert 0 °C ice to 0 °C water
The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
In this case, being 1 mol of water= 18 grams: Q= 10 g*= 3.333 kJ= 3,333 J (being kJ=1,000 J)
- Heat required to raise the temperature of water from 0 °C to 60 °C
In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 and ΔT=60 C - (0 C)= 60 C
Replacing: Q= 10 g*4.18 *60 C and solving: Q=2,508 J
Finally, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
<u><em> The amount of heat to absorb is 6,261 J</em></u>
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