freezing point depreesion
Homework Answers
Step1 Let M be molar mass of solute ; Moles of solute =1.2463/M
Step2 ΔTf= 4.9 ; Kf= 9.8 C/m
Step3 Molality of solute =(1.2463/M)(1000/10.9303)
Step4 ΔTf = Kf xMolality ; 9.8x1.2463x1000/Mx 10.9303 =4.9
Step5 M= 228.04
The freezing point of a solution of 1.104 g of an unknown nonelectrolyte dissolved in 36.81...
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please help ! 26. The freezing point depression of a solution of 1.45g sample of an...
please help !
26. The freezing point depression of a solution of 1.45g sample of an unknown nonelectrolyte dissolved in 25.00 mL of benzene(density d=0.879 g/mL) is 1.28°C. Pure benzene has Kf value of 5.12°C/m. What is the molecular weight of the compound? (10 points)
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please help !
26. The freezing point depression of a solution of 1.45g sample of an unknown nonelectrolyte dissolved in 25.00 mL of benzene(density d=0.879 g/mL) is 1.28°C. Pure benzene has Kf value of 5.12°C/m. What is the molecular weight of the compound? (10 points)
A 1.20 g sample of an unknown compound is dissolved in 60.0 g of benzene. The resulting solution freezes at 4.92 degree C. Calculate the molecular weight of the unknown. Pure benzene freezes at 5.45 degree C\ K_f for benzene is 5.12 degree C/m
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