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miskamm [114]
3 years ago
14

What is the molality of a solution of water and kcl if the freezing point of the solution is –3mc030-1.jpgc?

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
4 0
We will use the expression for freezing point depression ∆Tf
     ∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is 
     ∆Tf = 0C - (-3°C) = 3°C 
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and  one Cl- ion:
     KCl → K+ + Cl- 
Therefore, the molality m of the solution can be calculated as 
     3 = 2 * 1.86 * m
     m = 3 / (2 * 1.86)
     m = 0.80 molal
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Answer:

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Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

Thus, the atom percent turns out:

\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

Best regards.

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