1 (a). Based on the T values you obtained, plus information in the Pre -Lab exer

Question

1 (a). Based on the T values you obtained, plus information in the Pre -Lab exercise, predict the relative size of the T value for methanol. Compare its hydrogen-bonding capability and molecular weight to those of ethanol, 1-propanol, water, and acetone.   It is not important that you predict the exact T value; simply estimate a logical value that is higher, lower, or between the previous T values. (e.g. > 4°C or < 8°C). Record your predicted T, and then explain how you arrived at your answer.

(b). Do the same for hexane. Compare its hydrogen-bonding capability and molecular weight to
those of pentane, heptane, and octane. It is not important that you predict the exact T value; simply estimate a logical value that is higher, lower, or between the previous T values. (e.g. > 4°C or < 8°C). Record your predicted T, and then explain how you arrived at your answer.

For Methanol Delta T = -9.40 deg C

Pentane = -15.0 deg C

Heptane = -13.0 deg C

Octane = -6.20 deg C

Hexane = -13.3 deg C

Water = -5.40 deg C

Acetone = -19.30 deg C

Isopropyl aclohol = -5.50 deg C

Explanation / Answer

The magnitude of the dT value for any compound is determined by the following two factors :

Presence of H-bonds and molecular weight

Generally if intermolecular forces are strong the dT values are less

If H-bonding is present,there will be a small dT value. Absence of H-bonding leads to a greater dT value. This is the reason that as we proceed from water to methanol to isopropyl alcohol to acetone, the dT values increase because the intermolecular forces (H-bonding) decreases

If molecular weight is high, molecular size increases which increases the intermolecular interaction forces, so the dT value decreases. This is the reason that as we proceed from pentane to hexane to heptane to octane, the dT values decrease because size goes on increasing.

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