Achemistry student is given 600. ml of a clear aqueous solution at 43. ° c . he is told an unknown amount of a certain compound x is dissolved in the solution. the student allows the solution to cool to 23. ° c . at that point, the student sees that a precipitate has formed. he pours off the remaining liquid solution, throws away the precipitate, and evaporates the water from the remaining liquid solution under vacuum. more precipitate forms. the student washes, dries and weighs the additional precipitate. it weighs 0.054 kg . using only the information above, can you calculate the solubility of x in water at 23. ° c ? yes no if you said yes, calculate it. be sure your answer has a unit symbol and the right number of significant digits.

answer: - first one is the correct answer. .

explanations: - potassium sulfide is and cobalt(ii) chloride is .

charge on k is +1, charge on s is -2, charge on co is +2 and charge on cl is -1.

in a double replacement reaction the ion exchange takes place. in general the reaction looks like:

if one of the product is insoluble as per solubility rules then a precipitate of it is formed.

when we mix the aqueous solutions of potassium sulfide and cobalt(ii) chloride then chloride ion goes with potassium ion and sulfide ion goes with cobalt ion and the balanced equation is written as:

k and cl has equal charges, similarly co and s has equal charges and hence no subscripts are used for the product side.

as per the solubility rules, cos is insoluble and so it's precipitate is formed.

looking at the given choices, first choice is correct.