A) \[-270.90\text{ }K\]
B) \[270.90\text{ }K\]
C) \[273\text{ }K\]
D) \[274.15\text{ }K\]
E) \[-\text{ }274.15\text{ }K\]
Correct Answer: B
Solution :
\[\Delta {{T}_{f}}=\frac{{{k}_{f}}\times {{w}_{B}}\times 1000}{{{w}_{A}}\times {{w}_{B}}}\] \[{{w}_{A}}=600g,{{w}_{B}}=45g,{{k}_{f}}=1.86K\,kg\,mo{{l}^{-1}}\] \[{{M}_{B}}=62g\,mo{{l}^{-1}}\] \[\therefore \]\[\Delta {{T}_{f}}=\frac{1.86\times 45\times 1000}{600\times 62}=2.25\,K\] Hence, freezing point of aqueous solution \[=273.15-2.25\] \[=270.90\text{ }K\]
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