If the enthalpy of vaporization is approximat | vedclass

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(A) The ebullioscopic constant $(K_b)$ is given by the formula: $K_b = \frac{R \cdot M_1 \cdot T_b^2}{1000 \cdot \Delta H_{vap}}$.Here,$R$ is the gas

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Naphthalene

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(A) The ebullioscopic constant $(K_b)$ is given by the formula: $K_b = \frac{R \cdot M_1 \cdot T_b^2}{1000 \cdot \Delta H_{vap}}$.Here,$R$ is the gas constant,$M_1$ is the molar mass of the solvent,$T_b$ is the boiling point of the solvent,and $\Delta H_{vap}$ is the enthalpy of vaporization.Given that $\Delta H_{vap}$ is approximately the same for all,$K_b$ is directly proportional to $M_1 \cdot T_b^2$.Comparing the molar masses $(M_1)$ and boiling points $(T_b)$:$1$. Naphthalene $(C_{10}H_8)$: $M_1 = 128 \ g/mol$,$T_b = 491 \ K$$2$. Toluene $(C_7H_8)$: $M_1 = 92 \ g/mol$,$T_b = 384 \ K$$3$. Benzene $(C_6H_6)$: $M_1 = 78 \ g/mol$,$T_b = 353 \ K$$4$. Water $(H_2O)$: $M_1 = 18 \ g/mol$,$T_b = 373 \ K$Calculating the product $M_1 \cdot T_b^2$ shows that Naphthalene has the highest value due to its high molar mass and high boiling point.

If the enthalpy of vaporization is approximately the same,for which of the following will the value of the ebullioscopic constant $(K_b)$ be maximum?

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