Based on this image, What was the least precise measurement in the experiment? H

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Based on this image, What was the least precise measurement in the experiment? How does this limit your significant digits?

Beater Cse) 39 600 Molar Mass by Freezing Point Depression contined Mass Data Table Trial 2 Trial 1 Mass ofempty test tube #1. g Mass of test tube #1 plus BHT g Mass of BHT 25 013g Mass of weighing Mass of cetyl alcohol, g Mass ofempty test tube #2, g Mass of test tube #2 plus BHT, g Mass of BHT g Mass of unknown, g Cooling Data Table 10% 41e 49°C Temperature in C BHT+ Cetyl Alcohol Time, in Seconds BHT +Unknown Pure BHT 2 4O 3 630 76 13 220 280 300 Calculation Table BHT BHT + Cetyl Alcohol BHT+ Unknown Freezing point, C Molar mass, g/mole Percent error, molar mass 30 No prt of this material may be reprodoced or transmied in any form or by any means, electronic or mc without Catalog No. AP7948, from Flina Scientilic, Inc and only by mtans of paper copies ora from Fin Scientific Ine Student pages may be shared only by the instructor who purchased Molair Mass by Freeting Poiat Depesion

Explanation / Answer

starting from BHT data the posible intersection of the data shows the freezing point T= 69°C on the other hand in mixture of BHT+ cetyl alcohol (T = 60 °C) and mixture of BHT+ unknown (T= 66°C) from curve intersection plot.

now we first calculate the delta T i.e.

delta T = 60°C - 69°C = -9°C freezing point depression in BHT + Cetyl alcohol mixure

delta T = 66°C - 69°C = -3°C freezing point depression in BHT + unkown mixture

molecular weight of cetyl alcohol is = 242.45 g/mol

number of moles of cetyl alcohol in mixture is = 6.079 / 242.45 moles = 0.0251 moles

molality of Cetyl alcohol in mixture is = 0.0251

BHT is solvent here so its mass = 7.754 gm /1000 = 0.007754 kg

molality of solution is = 0.0251 / 0.007754 moles / kg = 3.237 moles/kg

Freezing point depression constant for BHT

Del T = Kf . m

9°C = Kf . 3.237

Kf = 2.780 deg. kg./moles

Now we can easily calculate these for unknown

Del T =3°C

Del T = Kf . m

m = Del T / Kf

m = 3 / 2.780 = 1.079 moles / kg

converting it into number of moles = 1.079 * 0.007177 = 0.007744 moles

by deviding the mass of the unknown we will get the molar mass of unknown i.e.

= 0.007744 * 1.059 = 136.75 g / mol

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