8.1grams of xso4.7H2O when heated at 90°c for 15minutes gave 4.9grams when heated further at 120°c. If x is 72,calculate the amount of water of crystallization in the intermediate and final dilute. (H=1;s=32,o=16,x=72).

The water of crystallization of the intermediate analyte is 5.

Seeing as the question does not provide the value for weight of the salt on further heating (at 120° C), you can apply the steps in this explanation to attain that value.

Explanation:

XSO4.7H2O → XSO4 + 7H2O

8.1g                        4.9g

8.1 - 4.9 = 3.9g of water evaporated

The ration of water to salt is:

                XSO4                :         H2O    

grams :        4.9                             3.2

moles:  72 + 32 + (16×2)=136               (1×2)+ 16= 18

       0.036                                      0.178 (divide both sides by smallest value)

        1                                               4.93

∴ water of crystallization of intermediate is 5.

answer: mercury

explanation:

the answer would be a