Question

When 282 gr of glycine are dissolved in 950 gr of a certain mystery liquid, the freezing point of the solution is 8.2 C less than the freezing point of pure. Calculate the mass of iron(III) chloride that must be dissolved in the same mass of to produce the same depression in freezing point. The van't Hoff factor for iron(III) chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.

Answer 1

`mass_(FeCl_3)=1.5x10^2gFeCl_3`

Step-by-step explanation:

Hello,

In this case, by using the given data for glycine, one computes the freezing point constant of the mystery liquid as shown below, considering the molality of the glycine and its van't Hoff factor equal to the unity:

`\Delta T=i*Kf*m_(Glyc)\\\\Kf=(\Delta T)/(i*m_(glyc)) =(8.2^oC)/(1*(282gGlyc)/(950gX)*(1molGlyc)/(75.07gGlyc)*(1000gX)/(1kgX) ) \\\\Kf=2.1^oC/m`

Now, as we are looking for the mass of iron(III) chloride at the same conditions of the aforesaid case, at first, one solves for the molarity of such compound considering that its theoretical van't Hoff factor is 4 as follows:

`m_(FeCl_3)=(\Delta T)/(i*Kf) =(8.2^oC)/(4*2.1^oC/m_(FeCl_3)) =0.98m`

Now, one obtains the requested mass via:

`mass_(FeCl_3)=0.98(molFeCl_3)/(kgX)*0.95kgX*(162.35gFeCl_3)/(1molFeCl_3)  \\\\mass_(FeCl_3)=151.1gFeCl_3\\mass_(FeCl_3)=1.5x10^2gFeCl_3`

Best regards.

Answer 2

Answer:
`1.4* 10^2g`

Step-by-step explanation:

Depression in freezing point is given by:

`\Delta T_f=i* K_f* m`

`\Delta T_f=T_f^0-T_f=8.2^0C` = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like glycine)

`K_f` = freezing point constant = ?

m= molality

`\Delta T_f=i* K_f* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}`

Weight of solvent = 950 g = 0.95 kg

Molar mass of glycine = 75.07 g/mol

Mass of glycine added = 282 g

`8.2=1* K_f* (282g)/(75.07 g/mol* 0.95kg)`

`K_f=2.2^0C/m`

Thus freezing point constant is
`2.2^0C/m`

2)
`\Delta T_f=i* K_f* m`

`\Delta T_f=T_f^0-T_f=8.2^0C` = Depression in freezing point

i= vant hoff factor = 4 (for
`FeCl_3`)

`K_f` = freezing point constant =
`2.2C/m`

m= molality

`\Delta T_f=i* K_f* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}`

Weight of solvent = 950 g = 0.95 kg

Molar mass of
`FeCl_3` = 162.2 g/mol

Mass of
`FeCl_3` added = ?

`8.2=4* 2.2* (xg)/(162.2 g/mol* 0.95kg)`

`x=1.4* 10^2g`

Thus mass of iron(III) chloride that must be dissolved in the same mass of to produce the same depression in freezing point is
`1.4* 10^2g`