1 mole glucose is added to 1l of water

Asked by admin @ 20/08/2021 in Chemistry viewed by 300 People

1 mole of glucose is added to 1L of water k=0.512K kg /mol calculate the boiling point of the solution?

Answered by admin @ 20/08/2021

Answer : The boiling point of the solution is, 372.48 K

Solution : Given,

Boiling point constant = 0.512 K kg/mole

Volume of solvent, water = 1 L = 1000 ml

Density of water = 0.9982 g/ml

Now first we have to calculate the mass of solvent, water.

$Density=\frac{Mass}{Volume}$

$Mass=0.9982g/ml\times 1000ml=998.2g$

Now we have to calculate the boiling point of the solution.

Formula used for Elevation in boiling point :

$\Delta T_b=k_b\times m$

or,

$T_{solvent}-T_{solution}=k_b\times \frac{\text{ Moles of solute}\times 1000}{\text{ Mass of solvent in Kg}}$

where,

$T_b$ = change in boiling point

$k_b$ = boiling point constant

m = molality

Temperature of solvent = 373 K

Now put all the given values in this formula, we get

$373K-T_{solution}=(0.512Kkg/mole)\times \frac{1moles\times 1000}{998.2kg}$

$T_{solution}=372.48K$

Therefore, the boiling point of the solution is, 372.48 K

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