Part A A solution of water (Kf=1.86 C/m) and glucose freezes at 4.35 C. What is

Question

Part A

A solution of water (Kf=1.86 C/m) and glucose freezes at 4.35 C. What is the molal concentration of glucose in this solution? Assume that the freezing point of pure water is 0.00 C. Express your answer to three significant figures and include the appropriate units.

Part B

A solution of water (Kb=0.512 C/m) and glucose boils at 103.56 C. What is the molal concentration of glucose in this solution? Assume that the boiling point of pure water is 100.00 C.

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

we have formula dT = i x Kf x m

where dT = difference in freezing points of solution and solvent = 0-(-4.35) = 4.35

i = vatoff factor = 1 for non dissociative compound like glucose

now we use formula to get m = molality

4.35 = 1 x 1.86 x m

m = 2.34

B) dT = i x kb x m is formula

dT = difference in boiling points of solution and solvent = 103.56-100 = 3.56

hence now 3.56 = 1 x 0.512 x m

m = molality = 6.95

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