Physics, 12.09.2019 19:30 Miloflippin9766

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

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Answer from: bskyeb14579

-0.45^0C

Explanation:-

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(0-T_f)^0C$ = Depression in freezing point

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

$KCl\rightarrow K^++Cl^-$

K_f = freezing point constant = 1.86^0C/m

m= molality

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

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

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

(0-T_f)^0C=0.45

T_f=-0.45^0C

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

Answer from: dnjames01

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

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

•2. mole fraction

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

$\large {\boxed {\bold {Xa = \frac {na} {na + nb}}}$

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

$\displaystyle \Delta T_f=K_f.m.i$

---> KCl is an electrolyte solution

$\Delta T_f=1.86.0.122.2$

$\displaystyle \Delta T_f=0.454$

freezing point of water = 0 °C

0.454=0-T~solution

T_f~solution=- 0.454 ^oC

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

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Part a calculate the freezing point of a solution containing 5.0 grams of kcl and 550.0 grams of wat...

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