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Propyl alcohol (C3H7OH) is a non-electrolyte. A 22.20-g sample of propyl alcohol is dissolved in 1.75 kg of water. If the boiling point elevation constant of water is 0.512 0C/m, what is the boiling point elevation of the solution?

Sagot :

The boiling point elevation of the solution : ΔTb = 0.108 °C

Further explanation  

Solution properties are the properties of a solution that don't depend on the type of solute but only on the concentration of the solute.  

The term is used in the Solution properties  

• 1. molal (m)

that is, the number of moles of solute in 1 kg of solvent  

[tex]\tt m=\dfrac{mol~solute}{mass~1~kg~solvent}[/tex]  

• 2. Boiling points  

Solutions from non-volatile substances have a higher boiling point and lower freezing points than the solvent  

ΔTb = Tb solution - Tb solvent  

ΔTb = boiling point elevation  

 ΔTb = Kb. m  

Kb = molal boiling point increase  

m = molal solution  

Given

mass alcohol = 22.2 g

mass water = 1.75 kg

Kb water = 0.512 °C/m

Required

ΔTb

Solution

mol alcohol :

= mass : MW

= 22.2 g : 60,0952 g/mol

= 0.369

molal of solution :

= 0.369 mol : 1.75 kg

= 0.211 m

The boiling point elevation :

ΔTb = 0.512 x 0.211

ΔTb = 0.108 °C

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