Please help with both , I keep getting -2.51 but the answer is actually (A) for

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Please help with both , I keep getting -2.51 but the answer is actually (A) for #20. I don't know how, so please include steps. And you have time can you explain why 21 is C? Thank you so much, any help is really appreciated. Please show all your work and explanation :) thank you :)

20. Calculate the freezing temperature of a solution made by mixing 63.672 g of CaCl2 (MW- 110.98) in 425.00 g of water. kf = 1.86 Cm-1 A. -7.53 °C B. -2.51 °C C. -10.09 °C D. -1.07 °C E. -3.20 °C 21. (Challenge Question) A reaction is run at 25°C and has an activation energy of 46kkJ/mol. Calculate the temperature at which you would need to run the reaction in order to increase the rte constant by a factor of change factor to 3 and Activation energy is 42.5 A. -4.73 °C B. 25.20 °C C. 45.39 °C D. 0.0031°C E. 320.9 °C

Explanation / Answer

1. Freezin point depresssion = iKf*m

m= molality= moles/ kg of solvent = 63.672/110.98/ 0.425 =1.35m

i=3 ( since there one Ca+2 and 2 Cl- ions)

Kf= 1.86c/m

freezing point depression = 3*1.35*1.86 =7.53 deg.c

Freezing point = 0-7.53= -7.53 deg.c

21.

from ln(K2/K1)= (Ea/R)*(1/T1-1/T2)

ln3 = (42.5*1000/8.314)*(1/298.15-1/T2)

0.000215= 1/298.15- 1/T2

1/T2= 1/298.15-0.000215=0.003139

T2= 1/0.003139=318.5 K (closest answer is 320.9)

=

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