m=(mass solute)(1mol solute/molar mass) / (mass of solvent) huuce any vomiting i
ID: 566087 • Letter: M
Question
m=(mass solute)(1mol solute/molar mass) / (mass of solvent)huuce any vomiting if ingested, and along with being inhaled bring the person to a place with fresh air and call a medical professional. The results for the experiment are listed in the table below. Based off this data, it is easily observed that the boiling point of the ethanol is raised each time more of the Urea is added. The more Urea in the solution, the more the boiling point elevates. Solution (M) Tb AT Mass (g) of Urea 75.0° C N/A 2.5° C 3.6° C 4.4° C 3.0° C Pure Ethanol N/A 0.5949 g 0.5848 g 0.5847 g 0.5 M 78.6° C 79.4° C 78.0° C 1.0 M 1.5 M Unknown 0.5923 g used when mixing the solutions. 0.592 g of Urea was the exact amount needed, but all the experimental rates ran a little high. The higher oncentration of solute would of raised the boiling point more than the experiment procedure led for. One other source of error was how the UJrea was originally dissobed Inctend ot One source of error is the actual masses of Urea
Explanation / Answer
Molarity = moles of solute/volume of solution (L)
moles of solute = molarity*volume of solution(L)
Molality = moles of solute/mass of solvent(kg)
= molarity*volume of solution(L)/mass of solvent(kg)
You know the molarities of the solution.
You need to know the total volume of solution.
And to calculate mass of solvent, you should know that how much (volume) solvent you added and then by using the density of ethanol (0.789 g/ml), you can convert it to mass of ethanol used. Then put all the values in the above formula, and you can convert molarity into molality.
For example for 0.5 M urea, you added 0.5949 g.
Moles of urea = mass/molar mass = 0.5949/60 = 0.0099 mol
0.5 M = 0.0099/Volume
volume = 0.0099/0.5 = 0.0198 L = 19.8 ml
Total volume of solution = 19.8 ml
As urea is a solid and 0.5949 g of urea was mixed in solvent to get 19.8 ml, we can assume that 19.8 ml of ethanol was added.
density of ethanol = 0.789 g/ml
mass of solvent = volume*density = 19.8*0.789 = 15.622 g = 0.0156 kg
Put all these values in the molality formula:
Molality = molarity*volume of solution(L)/mass of solvent(kg)
= 0.5*0.0198/0.0156 = 0.635 m
molality = 0.635 m
Similarly you can calculate for other molarities.
But if the mass of solute added is large, then this formula can not be used.
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