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Hey guys I did a lab and I have to answer the questions relating to the lab. I did question can someone please check if these a...Asked by Saira
Hey guys I did a lab and I have to answer the questions relating to the lab. I did question can someone please check if these are accurate
1. Calculation to Determine the molecular weight of unknown substance
Mass of unknown used: 2.0232 g
Mass of water used: 50.0 g (0.05kg)
Kf = 1.86°C kg/mole
Experimentally determined freezing point of unknown: -0.5°C
Initial Temperature = 9.20 C (had made a mistake)
Delta T= (-0.5) - 9.20
= -9.70
Solution = delta Tf/-Kf
= -9.70°C/-1.86°C kg/mole
= 5.2515 m (molality)
Molar mass = 2.0232g / 0.05 kg * 5.215 m
= 7.76 g/mol
I'm not sure how to do part B
2. Calculations to determine i for KCl solution:
GIVEN:
Delta t= t2-t1
= 0.00 -23.90
experimentally determined freezing point= 0.00
Molality KCl= 0.1
Kf= 1.86
1. Calculation to Determine the molecular weight of unknown substance
Mass of unknown used: 2.0232 g
Mass of water used: 50.0 g (0.05kg)
Kf = 1.86°C kg/mole
Experimentally determined freezing point of unknown: -0.5°C
Initial Temperature = 9.20 C (had made a mistake)
Delta T= (-0.5) - 9.20
= -9.70
Solution = delta Tf/-Kf
= -9.70°C/-1.86°C kg/mole
= 5.2515 m (molality)
Molar mass = 2.0232g / 0.05 kg * 5.215 m
= 7.76 g/mol
I'm not sure how to do part B
2. Calculations to determine i for KCl solution:
GIVEN:
Delta t= t2-t1
= 0.00 -23.90
experimentally determined freezing point= 0.00
Molality KCl= 0.1
Kf= 1.86
Answers
Answered by
bobpursley
The part
Solution=deltaTf/Kf
what is this? What is deltaTf? What was the initial freezing point of the solvent? If it was water, it was 0C. So deltatTf would be .5C
So that makes no sense to me.
Solution=deltaTf/Kf
what is this? What is deltaTf? What was the initial freezing point of the solvent? If it was water, it was 0C. So deltatTf would be .5C
So that makes no sense to me.
Answered by
Saira
So its
deltaTf = Kf * m
m= 9.70°C/-1.86°C kg/mole
= 5.2515 m (molality)
What i'm not sure about is, if
Delta Tf= in this equain is tfinal-t initial
or its the freezing point as ypu mentioned .5C
deltaTf = Kf * m
m= 9.70°C/-1.86°C kg/mole
= 5.2515 m (molality)
What i'm not sure about is, if
Delta Tf= in this equain is tfinal-t initial
or its the freezing point as ypu mentioned .5C
Answered by
bobpursley
Delta tf is the CHANGE in freezing point. The initial freezing point was not 9C. Water freezes at 0C, and then when you add a solute, the freezing point is depressed some.
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