A solution is prepared by dissolving 24.4g of C6H12O6 (non-electrolyte) in 0.500

Question

A solution is prepared by dissolving 24.4g of C6H12O6 (non-electrolyte) in 0.500 kg of ethanol. The boiling point of ethanol is 78.4 C and the boiling point elevation constant K(BP) is 1.22 (C•kg)/mole. What is the boiling point of the prepared solution? A solution is prepared by dissolving 24.4g of C6H12O6 (non-electrolyte) in 0.500 kg of ethanol. The boiling point of ethanol is 78.4 C and the boiling point elevation constant K(BP) is 1.22 (C•kg)/mole. What is the boiling point of the prepared solution?

Explanation / Answer

no of mol of sucrose(C6H12O6) = 24.4/180 = 0.135 mol

molality(m) = n*1/W

             = 0.135*(1/0.5)

             = 0.27 mol/kg

DTb = i*kb*m

(x-100) = 1*1.22*0.27

x = boiling point of solution = 100.33 C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.