A) \[16.02\text{ }g\]
B) \[1.602\text{ }g\]
C) \[0.1602\,g\]
D) \[160.2\text{ }g\]
Correct Answer: D
Solution :
\[KCl\xrightarrow{{}}{{K}^{+}}+C{{l}^{-}}\] 2 mole of \[KCl\] will give 2 mole particles and \[i=2\] \[\Delta T=i{{K}_{f}}m\] \[\Delta T=0-(-8.0)={{8}^{o}}C,\,i=2\] \[{{K}_{f}}={{1.86}^{o}}C\,kg\,mo{{l}^{-1}}\] \[8=2\times 1.86\times m\] \[m=2.15\,mol/kg\] Grams of \[KCl=2.15\times 74.5=160.2\,g/kg\] Hence \[160.2\text{ }g\] of \[KCl\]should be added to 1 kg of water.
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