Respuesta :
Explanation:
The given data is as follows.
Boiling point of water ([tex]T^{o}_{b}) = 100^{o}C[/tex] = (100 + 273) K = 323 K,
Boiling point of solution ([tex]T_{b}) = 101.24^{o}C[/tex] = (101.24 + 273) K = 374.24 K
Hence, change in temperature will be calculated as follows.
[tex]\Delta T_{b} = (T_{b} - T^{o}_{b})[/tex]
= 374.24 K - 323 K
= 1.24 K
As molality is defined as the moles of solute present in kg of solvent.
Molality = [tex]\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}[/tex]
Let molar mass of the solute is x grams.
Therefore, Molality = [tex]\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}[/tex]
m = [tex]\frac{288 g \times 1000}{x g \times 90}[/tex]
= [tex]\frac{3200}{x}[/tex]
As, [tex]\Delta T_{b} = k_{b} \times molality[/tex]
[tex]1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}[/tex]
x = [tex]\frac{0.512 ^{o}C/m \times 3200}{1.24}[/tex]
= 1321.29 g
This means that the molar mass of the given compound is 1321.29 g.
It is given that molecular formula is [tex]C_{n}H_{2n}O_{n}[/tex].
As, its empirical formula is [tex]CH_{2}O[/tex] and mass is 30 g/mol. Hence, calculate the value of n as follows.
n = [tex]\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]
= [tex]\frac{1321.29 g}{30 g/mol}[/tex]
= 44 mol
Thus, we can conclude that the formula of given material is [tex]C_{44}H_{88}O_{44}[/tex].
