Question

calculate the solubility of potassium bromide at 23 degrees C. HInt: Assume that the solubility increases by an equal amount for each degree between 20 degrees Cand 30 degrees C.

Answer 1

solubility of potassium bromide at 20 degrees celcius = 65.3

and at 30 degrees celcius = 70.7

now using the property of Lagrange interpolation formula,

we have to find the solubility at 23 degrees so

we have the value = L0f0 + L1f1

where L0 = ( 23 - 30)/ ( 20 - 30)

= 0.7

and f0 = 65.3

similarly L1 = ( 23 - 20 )/ ( 30 - 20)

= 0.3

and f1 = 70.7

so the value of solubility at 23 degrees = 0.7x65.3 + 0.3 x70.7

= 66.92 ( ans)