Answer:
Step-by-step explanation:
To determine the identity of the metal chloride, MCl, we need to calculate the mass of AgCl precipitate that was formed and compare it to the mass of the unknown chloride, MCl.
The mass of the filter paper plus AgCl precipitate is 2.23 g, and the mass of the filter paper alone is 0.80 g. Therefore, the mass of the AgCl precipitate is 2.23 g - 0.80 g = 1.43 g.
Since the reaction between AgNO3 and MCl resulted in the precipitation of AgCl, the mass of AgCl that was formed is equal to the mass of MCl that was present in the original solution. Therefore, we can compare the mass of AgCl precipitate that was formed to the mass of MCl that was provided to determine the identity of the metal chloride.
If we compare the masses, we can see that the mass of AgCl precipitate is 1.43 g, which is slightly less than the mass of MCl that was provided, which was 0.74 g. This suggests that the unknown metal chloride, MCl, is likely to be a compound with a lower molecular weight than AgCl.
One possibility is that MCl is a compound of a lighter metal, such as sodium or potassium. Sodium chloride (NaCl) and potassium chloride (KCl) have molecular weights of 58.44 g/mol and 74.55 g/mol, respectively, which are both lower than the molecular weight of AgCl (143.32 g/mol).
Therefore, it is likely that the unknown metal chloride, MCl, is either NaCl or KCl. To confirm the identity of the compound, additional analysis, such as a qualitative chemical test or a spectroscopic analysis, may be needed.