18g of glucose is dissolved in …
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Dr. Kamlapati Bhatt 8 years, 4 months ago
Calculations :
Moles of glucose , C6 H12 O6
= Mass of glucose / malar mass of glucose
= {tex}\frac{18}{180}{/tex} mol
= 0.1 mol
Number of Kg of solvent
= 1 kg ........(given)
{tex}\therefore{/tex}molality (m) of glucose solution
= 0.1 mol kg-1
For water , change in boiling point is given by the expression ,
{tex}\Delta T{/tex}b = Kb x m
where , Kb is Molal boiling point elevation constant is known as
= 0.52 K kg mol-1
and m represents molality
= 0.1 mol kg-1 ......calculated as above
So , substituting the values in above expression we get ,
{tex}\Delta Tb = 0.52kg mol^-1 X 0.1 mol kg^-1{/tex}
= 0.052 K
Since water boils at 373.15 K at 1.013 bar pressure , therefore the boiling point of solution will be
= ( 373.15 + 0.052 ) K
=373.202 K
0Thank You