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Q.

When 9.45 g of ClCH2COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of ClCH2COOH is  x×103.The value of x is ________. (Rounded off to the nearest integer).

KfH2O=1.86Kkgmol1          

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Detailed Solution

Moles of ClCH2COOH:

moles=massmolar mass moles=9.45 g94.5 g/mol=0.1 mol

          ClCH2COOHClCH2COO-+ H+ initial            1                      0                    0 At eq.        C1-α                              i=1+α

ΔTf=i×kf×m0.5=(1+α)×(1.86)×9.45×100094.5×5000.5=(1+α)×1.86×15(1+α)=0.51.86α=0.34

Dissociation constant (Ka) for ClCH2COOH:

Ka=.C(1-α) Ka=0.2×0.3421-0.34 Ka=35x×10-3

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When 9.45 g of ClCH2COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of ClCH2COOH is  x×10−3.The value of x is ________. (Rounded off to the nearest integer).KfH2O=1.86K kg mol−1