and grounds supervisor has asked you to evaluate alternative 1he Uhls for use on
ID: 549782 • Letter: A
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and grounds supervisor has asked you to evaluate alternative 1he Uhls for use on dis n the figre bedo amongst he goup and, on campus to melt the snow and ice this winter. als for use on sidewalks cam vide the compounds in the figure below from this figure, determine: n The molecular weight of each compound a. c. The maximum freezing point lowering that can be achieved for each compound ese Reconvene as a group and compare your results. d. Which compound is capable of lowering the freezing point the most? ls the order the same as the solubility? Why or why not? Generalize the trends you see in the data. e. Are there any compounds that you can eliminate as suitable materials for other reasons? 100 30 1 K20/207 0 80 KCI03 30 10 Ce2(304)3 100 0 8 Temperature (degrees Celsius) 0 10 20 30 40 50 60Explanation / Answer
1) The molar mass of the substances are
CaCl2 = 40 + 2x35.5= 111g/mole
NaNO3= 23+14+3*16= 85g/mole
KNO3= 101.1 g/mole, lead nitrate = 331.2 g/mole, K2Cr2O7= 294.1, KCl= 74.55 g/mole KClO3= 122.55 g/mole
NaCl= 58.44 g/mole and finally for calcium sulphate= 136.14
To find the maximum molality we need to check the solubility at 0oC. Because we are going to use them as antifreeze.
So for
1) For NaNO3= 73g/100g of water so molality = 73g/85gmole-1 * 0.1Kg= 8.588 and Freezing point change = 2*8.588*Kf( because NaCl will give two ions)=17.176*Kf
2) For CaCl2= 60g/100g of water Molality = 60g/111gmol-1x0.1= 5.4 and Freezing point change = 3*5.4*Kf=16.2*Kf
3) For Pb(NO3)2= 40g/100g of water: Molality 40/331.2*0.1= 1.2 moleKg-1 and freezing point change = 3x1.2xKf= 3.6Kf ( because lead nitrate will give three ions.
4) For KCl it is 28g/100g of water: Molality= 28/74.55x0.1 = 3.78 and freezing point change= 2x3.78x Kf= 4.56Kf (because KCl will give two ions)
5) For NaCl it is 40g/100g of water: Molality= 40/58.44*0.1 = 6.84 and freezing point change = 2*6.84*Kf= 13.68 Kf
The molality for other substances also can be found in the same way. Their solubilities are very less and hence the freezing point difference will be very less and negligible. By comparing we can say that NaNO3 is effective to remove the snow and the data can be interpretted from the solubility itself at 0oC.
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