Question

The molal boiling point constant of water is 0.53 c when 2 mole of glucose are dissolved in 4000 gm of water ,the solution will boil at

Answer 1

Answer: The solution will boil at $0.265^0C$

Explanation:- Formula used for Elevation in boiling point :

$\Delta T_b=k_b\times m$

or,

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

where,

$T_b$ = elevation in boiling point  

$k_b$ = boiling point constant  = 0.53

m = molality

$T_{solution}-0^0C=0.53\times \frac{2\times 1000}{4000g}$

$T_{solution}-0^0C=0.265$

$T_{solution}=0.265^0C$

Answer 2

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