The boiling point of water H 2 O is 100.0 °C at 1 atmosphere. A nonvolatile, non

Question

The boiling point of water H2O is 100.0°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is antifreeze (ethylene glycol).

How many grams of antifreeze, CH2OHCH2OH (62.10 g/mol), must be dissolved in 209.0 grams of water to raise the boiling point by 0.500 °C ?

g antifreeze??

The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is TNT (trinitrotoluene).

If 10.84 grams of TNT, C7H5N3O6 (227.1 g/mol), are dissolved in 236.1 grams of benzene ...

The molality of the solution is ?

The freezing point of the solution is °C?

Explanation / Answer

Increase in boiling point is calculated as

delta T = Kb x I x M

where M is molality

0.5 = 0.512 x 1 x M

M = 0.976

So molality needs to be 0.976 moles /Kg

SO in 209 g we must have 0.976 x 0.209 = 0.204 moles which is

0.204 x 62g/mol = 12.65 g of ethylene glycol needs to be dissolved in 209 g of water

10.84 g of TNT is 10.84/227.1g/mol = 0.0477 moles

In 236.1 g is 0.0477 x 1000/236.1 = 0.202 moles in 1000 g or 1 Kg

So molality is 0.202

delta T = Kf x i x M

delta T = 5.12 x 1 x 0.202

delta T = 1.03

So the new freezing point is 5.5-1.03 = 4.46 oC

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