Answer:
Approximately 10 g due to some experimental errors..
Explanation:
Hello,
In this case, since evaporation accounts for the removal of the solvent of a solution composed by a solvent and a solute via boiling it, we can see here that the solute is the salt and the solvent is the water and by performing an evaporation the 50 mL water need to be boiled; therefore at the end of the experiment, approximately 10 g of the salt are obtained as a solid product, but in some cases we could obtain more than such amount if the evaporation was not complete or less than it if decrepitation (spitting out of the solute due to the heat) occurs as the solid could "jump" out the recipient in which the experiment is carried out.
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Due to minor experimental flaws, the weight was estimated to be about 10 g.
In this particular instance, even though evaporation is defined as the removing the solvent from a facilitate successful of a solvent as well as a solute by boiling it, we would see that the substance is salt and the solvent is liquid, and that by conducting an evaporation.
50 mL water must be boiled; thus, at the end of the experiment, estimated 10 g of salt are acquired as a solid product; however, if the evaporative cooling was not comprehensive, we could acquire as much as this quantity.
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