A solution contains 5.36 g of an unknown electrolyte.?

Question

A solution contains 5.36 g of an unknown electrolyte. If 50.0 mL of water was used to make the solution, and if the solution freezes at -5.35°C, calculate the molecular weight of the compound and identify the unknown. Assume that the unknown is formed by elements from groups IA and VIIA.

Answer 1

Let's begin from the end :)
The electrolyte is formed by elements from group IA and VIIA. That means it is a binary compound with a cation that has charge +1 and an anion with -1.
The depression of the freezing point is

ΔT=i*Kf*m

where i is the van't Hoff coefficient and for strong electrolytes it is equal to the number of ions that are produced by the dissociation of the electrolyte according to its molecular formula. We said that the charges of the ions are +1 and -1, so for each cation we have 1 anion and thus from each electrolyte "molecule" we get only 2 ions (1 anion and 1 cation). Thus i=2.

m is the molality (moles of solute per kg of solvent).
mole=mass/MW, so m=1000*mass/(MW*G) where G is the mass of solvent expressed in grams.

so the equation becomes

ΔT=i*Kf*1000*mass/ (MW*G)

Let's assume that the density of water is exactly 1 g/mL and therefore 50.0 mL are 50.0 g of water.
For water Kf= 1.86
The freezing point of pure water is 0 so

0-(-5.35)= 2*1.86*1000* 5.36 /(MW*50) =>
MW =2*1.86*1000* 5.36 /(50*5.35)=74.54.

The only combination of such elements that gives this MW is KCl (74.55 or 74.5 depending on the level of accuracy you use for the atomic weights)