Respuesta :
First determine the molality of the solution. m = (w solute)/[(mm solute)(kg solvent)] -- mm means molar mass. So,
m = (14.7 g)/[(180 g/mol)(0.150 kg) =0.544 mol/kg = 0.544 m
Change in boiling point (∆t) = m x Kbp = (0.544 m)(0.512 °C/m) = 0.28°C
The solution will boil at 100.28°C at 1 atm.
the density if water is 1.00 g/ml and Kb= 0.512 C/m
The change in the boiling point of the solution has been 0.278 [tex]\rm ^\circ C[/tex]. The density of water at STP has been 1g/ml.
The change in the boiling point can be determined by:
Change in boiling point = molality × kb
The molality of the solution has been:
Molality = [tex]\rm \dfrac{weight\;of\;solute}{molecular\;weight\;of\;solute}\;\times\; \dfrac{1000}{Mass\;of\;solvent}[/tex]
Mass of solute = 14.7 g
Molecular mass solute = 180 g/mol
Mass of solvent = 150 grams
Substituting the values:
Molality = [tex]\rm \dfrac{14.7\;g}{180\;g/mol}\;\times\; \dfrac{1000}{150\;g}[/tex]
Molality = 0.544 m.
The kb of the water has been 0.512 [tex]\rm ^\circ C[/tex]/m
The change in the boiling point has been:
Change in boiling point = 0.544 m × 0.512 [tex]\rm ^\circ C[/tex]/m
Change in boiling point = 0.278 [tex]\rm ^\circ C[/tex]
The density of water at STP has been 1g/ml.
For more information about the change in boiling point, refer to the link:
https://brainly.com/question/10262489
