Question

When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing point of pure . Calculate the mass of ammonium chloride that must be dissolved in the same mass of to produce the same depression in freezing point. The van't Hoff factor for ammonium chloride in . Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.

Answer 1

Given question is incomplete. The complete question is as follows.

When 72.8 g of benzamide (
`C_(7)H_(7)NO`) are dissolved in 600 g of a certain mystery liquid X, the freezing point of the solution is
`6.90^(o)C` less than the freezing point of pure X. Calculate the mass of ammonium chloride
`(NH_(4)Cl)` that must be dissolved in the same mass of X to produce the same depression in freezing point. The van't Hoff factor i = 70 for ammonium chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.

Step-by-step explanation:

The given data is as follows.

Mass of solute (benzamide),
`w_(B)` = 72.8 g

Mass of solvent (X),
`w_(A)` = 600 g

`\Delta T_(f) 6.90^(o)C`

Molar mass of benzamide,
`M_{w_(B)}` = 121.14 g/mol

We know that,

`\Delta T_(f) = k_(f) * X * m` (for non-dissociating)

`6.90 = k_(f) * (72.8 * 1000)/(121.14 * 600)` ...... (1)

For other experiment, when
`NH_(4)Cl` is taken :

Mass of
`NH_(4)Cl`, (
`w_{NH_(4)Cl}`) = ?

Molar mass of
`NH_(4)Cl` = 53.491 g/mol

Mass of solvent (X) = 600 g

`\Delta T_(f) = 6.90^(o)C`

i = Van't Hoff factor = 1.70

As,
`\Delta T_(f) = i * k_(f) * m`

`6.90 = 1.70 * k_(f) * \frac{w_{NH_(4)Cl} * 1000}{53.491 * 600}` ........... (2)

Now, we will divide equation (1) by equation (2) as follows.

`w_{NH_(4)Cl} * 1 = (72.8 * 53.491)/(1.70 * 121.14)`

= 18.90 g

Therefore, we can conclude that the mass of ammonium chloride
`(NH_(4)Cl)` that must be dissolved in the same mass of X to produce the same depression in freezing point is 18.90 g.