Thank you for visiting At 293 K methanol has a vapor pressure of 97 7 Torr and ethanol has a vapor pressure of 44 6 Torr What would be. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Answer: 78.3 torr
Explanation:
According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.
[tex]p_1=x_1p_1^0[/tex] and [tex]p_2=x_2P_2^0[/tex]
where, x = mole fraction
in solution
[tex]p^0[/tex] = pressure in the pure state
According to Dalton's law, the total pressure is the sum of individual pressures.
[tex]p_{total}=p_1+p_2[/tex]
[tex]p_{total}=x_Ap_A^0+x_BP_B^0[/tex]
moles of ethanol= [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{80g}{46g/mol}=1.7moles[/tex]
moles of methanol= [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{97g}{32g/mol}=3.0moles[/tex]
Total moles = moles of ethanol + moles of methanol = 1.7 +3.0 = 4.7
[tex]x_{ethanol}=\frac{1.7}{4.7}=0.36[/tex],
[tex]x_{methanol}=1-x_{ethanol}=1-0.36=0.64[/tex],
[tex]p_{ethanol}^0=44.6torr[/tex]
[tex]p_{methanol}^0=97.7torr[/tex]
[tex]p_{total}=0.36\times 44.6+0.64\times 97.7=78.3torr[/tex]
Thus the vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 K is 78.3 torr.
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Rewritten by : Jeany
In a closed arrangement when the gases exert pressure at their consolidated stages then it is called vapour pressure.
The correct answer is:
Option D. 78.3 torr.
This can be calculated by:
According to Raoult's law:
[tex]\text{P}_{\text{solution}} & = X_{\text{solvent}} P_{\text{solvent}}[/tex]
Where,
x = mole fraction
Pâ°= pressure in the pure form
Now according to Dalton:
The total of the individual pressure would be equivalent to the total pressure.
[tex]\text{P}_{\text{total}} & = \text{P}_{1} +\text{ P}_{2}[/tex]
[tex]\text{P}_{\text{total}} & = \text{x}_{\text{A}}\text{P}^{0} _{\text{A}} + \text{x}_{\text{B}}\text{P}^{0} _{\text{B}}[/tex]
- Moles of ethanol can be calculated as:
[tex]\text{Moles} &= \dfrac{\text{Mass}}{\text{Molar mass}}[/tex]
[tex]\text{Moles} &= \dfrac{\text{80 g}}{\text{46 g/mol}} & = 1.7 \; \text{moles}[/tex]
- Moles of methanol can be calculated as:
[tex]\text{Moles} &= \dfrac{\text{Mass}}{\text{Molar mass}}\\[/tex]
[tex]\text{Moles} &= \dfrac{\text{97 g}}{\text{ 32 g/mol}} & = 3.0 \; \text{moles}[/tex]
- Total moles = Moles of Ethanol + Moles of Methanol
= 1.7+ 3.0
= 4.7 moles
[tex]\text{x}_{\text{ethanol}} = \dfrac{1.7} {4.7} = 0.36[/tex]
[tex]\text{x}_{\text{methanol}} = 1 - \text{x}_{\text{ethanol}} = 1- 0.36 = 0.64[/tex]
[tex]\text{P}^{0} _{\text{ethanol}} = 44.6 \text{torr}[/tex]
[tex]\text{P}^{0} _{\text{methanol}} = 97.7 \text{torr}[/tex]
[tex]\text{P}_{\text{total}} = 0.36 \; \times 44.6\; + 0.64\; \times 97.7 = 78.3 \;\text{torr}[/tex]
Therefore, the vapour pressure of a mix of 80 g of ethanol and 97 g of methanol is 78.3 torr.
To learn more about vapour pressure refer to the link:
https://brainly.com/question/2693029
