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A 0.1 molal solution of an acid is 4.5 % ionized. Calculate freezing point. (molecular weight of the acid is 300 ) . K_f=1.86 ~K ~mol ⁻¹ ~kg .

Options

  1. A-0.194^ C
  2. B2.00^ C
  3. C0^ C
  4. D-0.269^ C

Correct answer

A. -0.194^ C

Step-by-step solution

If acid is 4.5 % ionized then =0.045 . aligned & T_f= molality K_f=0.1 1.86=0.186 & T_ = T_f(1+ )=0.186(1+0.045)=0.194^ C aligned

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