How many grams of a sucrose ($mol\ wt. = 342$) should be dissolved in 100 g water in order to produce a solution with a $105.0℃$ difference between the freezing point and boiling temperature? $(k_{f} = 1.86\ C/m,\ k_{b} = 0.151℃)$

Question

Dr. Abubacker · Accepted Answer

(b)
Boiling point $\left( T_{b} ight) = 100 + \mathrm{\Delta}T_{b} = 100 + k_{b}m$
Freezing point $\left( T_{f} ight) = 0 - \mathrm{\Delta}T_{f} = - k_{f}m$$T_{b} - T_{f} = \left( 100 + k_{b}m ight) - ( - k_{f}m)$
$105 = 100 + 0.51m + 1.86m$
$2.37\ m = 5\ or\ \ m = \frac{5}{2.37} = 2.11$
$	herefore$ Weight of sucrose to be dissolved in 100 g water
$= \frac{2.11 	imes 342}{1000} 	imes 100 = 72g$

Q. How many grams of a sucrose ($mol\ wt. = 342$) should be dissolved in 100 g water in order to produce a solution with a $105.0℃$ difference between the freezing point and boiling temperature? $(k_{f} = 1.86\ C/m,\ k_{b} = 0.151℃)$

Explanation

Reviewed by Dr. Abubacker

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