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

Step-by-step 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 * 1000}}{\text{molar mass of solute * mass f solvent(g)}}`

Let molar mass of the solute is x grams.

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

m =
`(288 g * 1000)/(x g * 90)`

=
`(3200)/(x)`

As,
`\Delta T_(b) = k_(b) * molality`

`1.24 = 0.512 ^(o)C/m * (3200)/(x)`

x =
`(0.512 ^(o)C/m * 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}}`

=
`(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)`.