Question

One solution has a formula C (n) H (2n) O (n) If this material weighs 288 grams, dissolves in weight 90 grams, the solution will have a boiling point of 101.24 ° C. Find the formula
The molecules of this substance When determining the Kb value of water = 0.512 ° C / m and the atomic weight H = 1, C = 12 and O = 16.

Answer 1

Explanation:

The given data is as follows.

Boiling point of water ($T^{o}_{b}) = 100^{o}C$ = (100 + 273) K = 323 K,

Boiling point of solution ($T_{b}) = 101.24^{o}C$ = (101.24 + 273) K = 374.24 K

Hence, change in temperature will be calculated as follows.

              $\Delta T_{b} = (T_{b} - T^{o}_{b})$

                           = 374.24 K - 323 K

                           = 1.24 K

As molality is defined as the moles of solute present in kg of solvent.

            Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

Let molar mass of the solute is x grams.

Therefore,   Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

                        m = $\frac{288 g \times 1000}{x g \times 90}$              

                          = $\frac{3200}{x}$

As,    $\Delta T_{b} = k_{b} \times molality$

                 $1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}$

                       x = $\frac{0.512 ^{o}C/m \times 3200}{1.24}$

                          = 1321.29 g

This means that the molar mass of the given compound is 1321.29 g.

It is given that molecular formula is $C_{n}H_{2n}O_{n}$.

As, its empirical formula is $CH_{2}O$ and mass is 30 g/mol. Hence, calculate the value of n as follows.

                n = $\frac{\text{Molecular mass}}{\text{Empirical mass}}$

                   = $\frac{1321.29 g}{30 g/mol}$

                   = 44 mol

Thus, we can conclude that the formula of given material is $C_{44}H_{88}O_{44}$.