Question

calculate the boiling point of 1 molar aqueous solution of solute (molar mass =74.5 g mol-1). the density of solution is 1.04 g ml-1 and kb for water is 0. 52 k kg mol-1

Answer 1

Answer:

Explanation:

m - molality

M - molarity

d - density

MB - molar mass

Answer 2

Answer:

101.078C

Explanation:

• KCl's molar mass is 39+35.5=74.5 g/mol.
• As KCl entirely decomposed (KCL -> K+ + Cl-), the quantity of ions generated was equal to 2.
• Thus, the Vant Hoff factor is two.
• KCl solution mass equals 1000*1.04 = 1040g.
• Solvent mass equals 1040-74.5g, or 965.5g.
• Molality (m) is equal to 1 mol of solute divided by 0.9655 kg of solvent, or 1.0357 m.
• Tb=imkb equals 2*1.0357*0.52 = 1.078C.
• Thus, the solution's boiling point is 100 + 1.078 = 101.078C.
• The Van't Hoff factor sheds light on how solutes affect the collaberative characteristics of solutions. The letter I is used to represent it. The ratio of a substance's mass concentration to the concentration of the particles that are produced when it dissolves is known as the Van't Hoff factor.

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