Respuesta :
Given question is incomplete. The complete question is as follows.
When 72.8 g of benzamide ([tex]C_{7}H_{7}NO[/tex]) are dissolved in 600 g of a certain mystery liquid X, the freezing point of the solution is [tex]6.90^{o}C[/tex] less than the freezing point of pure X. Calculate the mass of ammonium chloride [tex](NH_{4}Cl)[/tex] 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.
Explanation:
The given data is as follows.
Mass of solute (benzamide), [tex]w_{B}[/tex] = 72.8 g
Mass of solvent (X), [tex]w_{A}[/tex] = 600 g
[tex]\Delta T_{f} 6.90^{o}C[/tex]
Molar mass of benzamide, [tex]M_{w_{B}}[/tex] = 121.14 g/mol
We know that,
[tex]\Delta T_{f} = k_{f} \times X \times m[/tex] (for non-dissociating)
[tex]6.90 = k_{f} \times \frac{72.8 \times 1000}{121.14 \times 600}[/tex] ...... (1)
For other experiment, when [tex]NH_{4}Cl[/tex] is taken :
Mass of [tex]NH_{4}Cl[/tex], ([tex]w_{NH_{4}Cl}[/tex]) = ?
Molar mass of [tex]NH_{4}Cl[/tex] = 53.491 g/mol
Mass of solvent (X) = 600 g
[tex]\Delta T_{f} = 6.90^{o}C[/tex]
i = Van't Hoff factor = 1.70
As, [tex]\Delta T_{f} = i \times k_{f} \times m[/tex]
[tex]6.90 = 1.70 \times k_{f} \times \frac{w_{NH_{4}Cl} \times 1000}{53.491 \times 600}[/tex] ........... (2)
Now, we will divide equation (1) by equation (2) as follows.
[tex]w_{NH_{4}Cl} \times 1 = \frac{72.8 \times 53.491}{1.70 \times 121.14}[/tex]
= 18.90 g
Therefore, we can conclude that the mass of ammonium chloride [tex](NH_{4}Cl)[/tex] that must be dissolved in the same mass of X to produce the same depression in freezing point is 18.90 g.
