a. 1,4332 g
b. 7.54~g
<h3>Further explanation</h3>
Given
Reaction
MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)
20 cm of 2.5 mol/dm^3 of MgCl2
20 cm of 2.5 g/dm^3 of MgCl2
Required
the mass of silver chloride - AgCl
Solution
a. mol MgCl2 :

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1
mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g
b. mol MgCl2 (MW=95.211 /mol):

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526
mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g
Composed of molecules formed by atoms of two or more different elements.
Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.