What is the boiling-point $\underline{\text{ change }}$ for a solution containing 0.432 mol of naphthalene (a nonvolatile,nonionizing compound)in 250.g of liquid benzene? (K_bp = 2.53°C/m for benzene) A) 4.37 °C B) 5.86 °C C) 0.273 °C D) 1.46 °C E) 1.093 °C

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The boiling-point change is given by the equation:$$\Delta T_b = K_{bp} 	imes m$$where:* $\Delta T_b$ is the boiling-point change in °C* $K_{bp}$ is the boiling-point elevation constant in °C/m* $m$ is the molality of the solution in mol/kgIn this problem, we have:* $K_{bp}$ = 2.53 °C/m* $m$ = 0.432 mol / 0.250 kg = 1.728 mol/kgSubstituting these values into the equation, we get:$$\Delta T_b = 2.53 °C/m 	imes 1.728 mol/kg = 4.37 °C$$Therefore, the boiling-point change is 4.37 °C.

What Is the Boiling-Point change ‾\underline{\text{ change }} change ​ For a Solution Containing 0

What Is the Boiling-Point $\underline{\text{ change }}$ For a Solution Containing 0