The boiling point elevation of an aqueous sucrose solution is found to be 0.39°C.What mass of sucrose (molar mass= 342.30 g/mol)would be needed to dissolve in 500.0 g of water? K_b (water)= 0.512°C/m. A) 261 g sucrose B) 528 g sucrose C) 762 g sucrose D) 223 g sucrose E) 130. g sucrose

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The boiling point elevation ($\Delta T_b$) is given by the formula $\Delta T_b = i \cdot K_b \cdot m$, where $i$ is the van't Hoff factor (which is 1 for sucrose, a non-electrolyte), $K_b$ is the ebullioscopic constant of the solvent (water in this case, 0.512°C/m), and $m$ is the molality of the solution. Given $\Delta T_b = 0.39°C$, we can solve for $m$ (molality):$$0.39°C = 1 \cdot 0.512°C/m \cdot m$$$$m = \frac{0.39}{0.512} \, \text{mol/kg}$$$$m = 0.7617 \, \text{mol/kg}$$Molality ($m$) is defined as moles of solute per kilogram of solvent. To find the mass of sucrose needed, we first find the moles of sucrose:$$0.7617 \, \text{mol/kg} \times 0.500 \, \text{kg} = 0.38085 \, \text{mol}$$Then, using the molar mass of sucrose (342.30 g/mol), we find the mass of sucrose:$$0.38085 \, \text{mol} \times 342.30 \, \text{g/mol} = 130.3 \, \text{g}$$Therefore, approximately 130 g of sucrose is needed.

The Boiling Point Elevation of an Aqueous Sucrose Solution Is