At 20^o C,benzene and toluene form an ideal s | vedclass

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(B) $1$. Calculate the moles of benzene $(C_6H_6)$: $n_{benzene} = \frac{78 \ g}{78 \ g/mol} = 1 \ mol$. $2$. Calculate the moles of toluene $(C_7H_8)

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$50$

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(B) $1$. Calculate the moles of benzene $(C_6H_6)$: $n_{benzene} = \frac{78 \ g}{78 \ g/mol} = 1 \ mol$. $2$. Calculate the moles of toluene $(C_7H_8)$: $n_{toluene} = \frac{46 \ g}{92 \ g/mol} = 0.5 \ mol$. $3$. Calculate the mole fraction of benzene $(X_{benzene})$: $X_{benzene} = \frac{n_{benzene}}{n_{benzene} + n_{toluene}} = \frac{1}{1 + 0.5} = \frac{1}{1.5} = \frac{2}{3}$. $4$. Calculate the partial vapor pressure of benzene $(P_{benzene})$ using Raoult's Law: $P_{benzene} = P^0_{benzene} 	imes X_{benzene} = 75 \ torr 	imes \frac{2}{3} = 50 \ torr$.

At $20^o C$,benzene and toluene form an ideal solution. The pure vapor pressure of benzene is $75 \ torr$ and that of toluene is $22 \ torr$. What is the partial vapor pressure of benzene in a solution containing $78 \ g$ of benzene and $46 \ g$ of toluene at $20^o C$?

Similar Questions

$A$ liquid mixture is an ideal solution, if
$(I)$ It obeys ideal gas equation
$(II)$ It obeys Raoult's law at all concentrations
$(III)$ Solute - solute, solute - solvent and solvent - solvent interactions are similar

In a mixture of $A$ and $B$ components,the solution shows positive deviation from Raoult's law when:

An ideal solution has two components $A$ and $B$. If $A$ is more volatile than $B$ and also $P_A^o > P_T$,then the correct relation between mole fraction of $A$ in liquid $(x)$ and vapour $(y)$ phase is

What is vapour pressure? And what is an ideal solution?

Assertion $A$: For an ideal solution formed by mixing liquids $P$ and $Q$, $\Delta_{mix} H = 0$ and $\Delta_{mix} V = 0$.
Reason $R$: No interactions occur between $P$ and $Q$.
In the light of the above statements, choose the most appropriate answer from the options given below:

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