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.
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
P⁰= 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