Question

45 g of ethylene glycol is mixed with 600 g of water. What is the freezing point of the solution? $ ({{k}_{f}}=1.86\text{ }K\text{ }kg\text{ }mo{{l}^{-1}}) $

• 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 $

Answer 1

Answer: (B)

$ \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 $