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Part A Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depression constant (Kf) for water is 1.86 °C/m. Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depression constant () for water is 1.86 °C/m. +0.23 °C -0.23 °C 1.23 °C -0.45 °C +0.45 °C

Respuesta :

Answer: [tex]-0.45^0C[/tex]

Explanation:-

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=T_f^0-T_f=(0-T_f)^0C[/tex] = Depression in freezing point

i= vant hoff factor = 2 (for electrolyte undergoing complete dissociation, i is equal to the number of ions produced)

[tex]KCl\rightarrow K^++Cl^-[/tex]

[tex]K_f[/tex] = freezing point constant = [tex]1.86^0C/m[/tex]

m= molality

[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

Weight of solvent (water)= 550 g = 0.55 kg  (1kg=1000g) 

Molar mass of solute (KCl) = 74.5 g/mol

Mass of solute (KCl) = 5.0 g

[tex](0-T_f)^0C=2\times 1.86\times \frac{5g}{74.5g/mol\times 0.55kg}[/tex]

[tex](0-T_f)^0C=0.45[/tex]

[tex]T_f=-0.45^0C[/tex]

Thus the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water is [tex]-0.45^0C[/tex]

ardni313

The freezing point of a solution : -0.45 °C

Further explanation  

Solution properties are the properties of a solution that don't depend on the type of solute but only on the concentration of the solute.  

Solution properties of electrolyte solutions differ from non-electrolyte solutions because electrolyte solutions contain a greater number of particles and break down into ions. So the Solution properties of electrolytes is greater than non-electrolytes

The term is used in the Solution properties

• 1. molal  

that is, the number of moles of solute in 1 kg of solvent  

[tex]\large {\boxed {\bold {m = mole. \frac {1000} {mass \: of \: solvent (in \: grams)}}}[/tex]

• 2. mole fraction  

the ratio of the number of moles of solute to the mole of solution  

[tex]\large {\boxed {\bold {Xa = \frac {na} {na + nb}}}[/tex]

a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depression constant () for water is 1.86  

• Step 1: determine molal  

molar mass KCl = 39 + 35.5 = 74.5  

mole KCl = mass: molar mass  

mole KCl = 5 gr: 74.5  

mole KCl = 0.067  

molal = m = 0.067 x (1000: 550 gr water)  

molal = 0.122  

• Step 2: determine the freezing point of the solution  

i = 1 + (n-1) a  

i = 1 + (2-1) 1  

i = 2  

[tex]\displaystyle \Delta T_f=K_f.m.i[/tex]

---> KCl is an electrolyte solution  

[tex]\Delta T_f=1.86.0.122.2[/tex]

[tex]\displaystyle \Delta T_f=0.454[/tex]

freezing point of water = 0 °C  

[tex]0.454=0-T~solution[/tex]

[tex]T_f~solution=- 0.454 ^oC[/tex]

Learn more  

Raoult's law  

https://brainly.com/question/10165688  

The vapor pressure of benzene  

https://brainly.com/question/11102916  

The freezing point of a solution  

https://brainly.com/question/8564755  

https://brainly.com/question/4593922  

https://brainly.com/question/1196173  

Keywords: Freezing Point Depression, Boiling Point Elevation, Solution Properties  

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