What is the boiling point of a solution of 0.1 mole of glucose in 200 mL of water? a) 100.06 degrees C b) 100.13 degrees C c) 100.26 degrees C d) 100.5 degrees C

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To solve for the new boiling point of the solution, we can use the expression:

`Delta T = m * K_(b)`

Where:

`Delta T = T_(t o t a l) - T_(solvent)`

`T_(solvent) = T_(H_2 O) = 100 ^0 C`

`m = molality = (mol es)/(kg solvent)`

`K_b = 0.512 for water`

 

for water:

`200 mL = 200 grams = 0.200 kg`

 

for molality:

`m = (0.100 mol es)/(0.200 kg)`

`m = 0.500`

 

 

`Delta T = m * K_(b)`

`Delta T = 0.512 * 0.500 = 0.256 ^0 C`

`T_(t o t a l) - T_(solvent) = 0.256 ^0 C`

`T_(t o t a l) = 0.256 + 100 = 100.256 = 100.26`

 

Answer: letter C

 

Sources:

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