Boiling-point elevation is a colligative property.
That means, the the boiling-point elevation depends on the molar content (fraction) of solute.
The dependency is ΔTb = Kb*m
Where ΔTb is the elevation in the boiling point, kb is the boiling constant, and m is the molality.
A solution of 6.00 g of Ca(NO3) in 30.0 g of water has 4 times the molal concentration of a solution of 3.00 g of Ca(NO3)2 in 60.0 g of water.:
(6.00g/molar mass) / 0.030kg = 200 /molar mass
(3.00g/molar mass) / 0.060kg = 50/molar mass
=> 200 / 50 = 4.
Then, given the direct proportion of the elevation of the boiling point with the molal concentration, the solution of 6.00 g of CaNO3 in 30 g of water will exhibit a greater boiling point elevation.
Or, what is the same, the solution with higher molality will have the higher boiling point.
From the periodic table:
molar mass of Sn = 118.71 grams
molar mass of F = 18.99 grams
This means that:
molar mass of SnF2 = 118.71 + 2(18.99) = 156.69 grams
Therefore, 156.69 grams of SnF2 contains 37.98 grams of F. To know the amount of F in 36.5 grams of the compound, we will simply do a cross multiplication as follows:
mass of F = (36.6 x 37.98) / 156.69 = 8.847 grams
I think it’s the inner membrane
