<span>There are five. Two atoms of aluminum, and three atoms of oxygen.There are 5</span>
Answer : The molecular weight of this compound is 891.10 g/mol
Explanation : Given,
Mass of compound = 12.70 g
Mass of ethanol = 216.5 g
Formula used :

where,
= change in freezing point
= temperature of pure ethanol = 
= temperature of solution = 
= freezing point constant of ethanol = 
i = van't hoff factor = 1 (for non-electrolyte)
m = molality
Now put all the given values in this formula, we get


Therefore, the molecular weight of this compound is 891.10 g/mol
Molar mass MgCO3 => 84.31 g/mol
1 mole MgCO3 ----------- 84.31 g
1.5 moles MgCO3-------- ??
1.5 x 84.31 / 1 => 126.465 g