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18 votes
What is the boiling point of a solution of 1.0 g of glycerin in 47.8 g of water?

User Nicknow
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2 Answers

16 votes
16 votes

Answer:

100.12*C

Step-by-step explanation:

Our solute is glycerin and the solvent is water.

1) Convert grams of the solute to moles:

n (moles)=Mass/Molar Mass => 1/92.1=0.01086 moles

2) Find molality from moles of solute and mass of solvent in Kg:

m (molality)=moles of solute/Kg solvent => 0.01086/0.0478=0.2272m

3) Finding the change boiling point (Bp) using molality times the Bp constant (the Bp constant of water is 0.512*C/m): 0.2272*0.512=0.116*C

4) Under 1 atm, the boiling point of water is exactly 100 and to find the new Bp, simply add the change Bp completed in step 3 with the boiling point of water: 100+0.116=100.116*C or 100.12*C

Hope this helps!

User Setsu
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14 votes
14 votes
100 degrees c is the answer
User Hughsk
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