Chemistry-
General
Easy

Question

What is the solubility product of C a F subscript 2 end subscript comma if its saturated solution contains 0.017 blank g of C a F subscript 2 end subscript per l i t r e?

  1. 1.44 cross times 10 to the power of negative 4 end exponent  
  2. 4.14 cross times 10 to the power of negative 11 end exponent  
  3. 4.14 cross times 10 to the power of negative 18 end exponent  
  4. 41.4 cross times 10 to the power of negative 24 end exponent  

The correct answer is: 4.14 cross times 10 to the power of negative 11 end exponent


    K subscript s p end subscript equals 4 s to the power of 3 end exponent
    A l s o comma blank s equals fraction numerator 0.017 over denominator 78 end fraction M
    therefore K subscript s p end subscript equals 4 cross times open parentheses fraction numerator 0.017 over denominator 78 end fraction close parentheses to the power of 3 end exponent equals 4.14 cross times 10 to the power of negative 11 end exponent

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