[spKelpDiver]

2.  The thermometer you use to measure the freezing temperature of the solvent actually
records a temperature that is 0.12°C too high.  Unfortunately you break this thermometer
and borrow a fellow student's thermometer, which reads 0.11°C too low. If the molality
of your solution is 0.30, calculate the contribution to the percentage error in your result
due to not correcting for the errors in the thermometer readings.

Given: Solvent = cyclohexane,  Kf = 20 °C/m

I just wanted to make sure my method was correct.

1.) First I solved for ΔT, measured by the borrowed thermometer, using the 0.30 molality and the freezing point constant of cyclohexane 20 °C/m 

ΔT= (.30m) x 20 °C/m = 6.0 °C

2.) Next I solve for what the original molality would have been if the temperature was measured with the broken thermometer.

6.0 °C is 0.11 °C too low, but the original (broken) thermometer is 0.12 °C too high. So original (broken) thermometer would have been 6.0 + 0.11 + .12 = 6.23 °C
 
Solving for m.... m = ΔT/Kf   =   6.23 °C / 20°C·m^-1  = 0.31 molality.

3.) Now I solve for percent error.. Theoretical = 0.30m Actual = 0.31m
theoretical - actual / actual x 100  = 0.30m - 0.31m / 0.31m   x 100 = 3.2% error.



Is this right?

[Vidya]

remember all %ages  are calculated on the true values .Theoretical value is the true value
so the formula is |theroretical -actual|/theroretical X 100%