Question

a solution of glycerol in water was prepared by dissolving some glycerol in 500g. of water.ts solution has a boiling point of 100.42. what mass of glycerol was disssolved to make ts solution

Answer 1

ΔTb = Kb x m 
100.42 -100 = 0.512 x m
m = 0.42/0.512  =  0.82 
no. of moles of glycerol/mass(in kg) of solvent(water) = 0.82
(x/92)/0.5  = 0.82
x/92 = 0.82 x 0.5 =   0.41

x = 92 x 0.41 =   37.72 g 

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Answer 2

Answer: 37.73 g

Explanation :

Weight of solvent (water)=500 g = 0.5 kg      (1 kg=1000 g)

Molar mass of glycerol $(C_3H_8O_3)$ = 92 g/mol

Mass of glycerol added = ? g

$\Delta T_b=K_b\times \frac{\text{mass of glycerol}}{\text{molar mass of glycerol}\times \text{weight of solvent in kg}}$

$\Delta T_b$ = elevation in boiling point

$K_b$ = boiling point constant =$0.512°C kg/mol$

$\Delta T_b=T_b-T_b^0=(100.42-100)^0C=0.42^0C$

$0.42^0C=0.512\frac{xg}{92 g/mol\times 0.5kg}$

$x=37.73g$

Thus the mass of glycerol added was 37.73 g