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damaskus [11]
3 years ago
15

The normal freezing point of a certain liquid Xis-7.30°C but when l02. g of iron(III) chloride (FeCl3) are dissolved in 650. g o

f Xthe solution freezes at -9.9°C instead. Use this information to calculate the molal freezing point depression constant Kf of X.
Chemistry
1 answer:
IRISSAK [1]3 years ago
7 0

Answer:

2.7 °C.kg/mol

Explanation:

Step 1: Calculate the freezing point depression (ΔT)

The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:

ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C

Step 2: Calculate the molality of the solution (b)

We will use the following expression.

b = mass of solute / molar mass of solute × kilograms of solvent

b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg

Step 3: Calculate the molal freezing point depression constant Kf of X

Freezing point depression is a colligative property. It can be calculated using the following expression.

ΔT = Kf × b

Kf = ΔT / b

Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol

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Chemistry an atom of strontium has at least four different isotopes. what is different between an isotope of 86sr and an isotope
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3 years ago
What does AuCl, Gold (I) Chloride, decompose into?
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6 0
3 years ago
A container of carbon dioxide has a volume of 240 mL at a temperature of 22°C. If the pressure remains constant, what is the vol
astraxan [27]

Answer:

Volume of the CO_{2} gas at 44°C is <u>258 ml.</u>

Explanation

here,

using Charles' law ,

\frac{V}{T} =\frac{v}{t}

where , V= initial volume          v= final volume

             T=initial temperature    t = final temperature

Given - pressure is constant ,

so , putting the values -

V= 240ml

T= 22 + 273K = 295K                      (since converting celsius into kelvin that

                                                         is +273K )

v =?

t = 44+ 273K = 317K

Now , putting the given values in charles' law ,

\frac{240ml}{295K} =\frac{v}{317K}

240ml x317K = v x 295K     (through cross multiplication )

v =\frac{240ml\times317K}{295K}

= 258ml .

thus ,<u> the volume of carbon dioxide in a container at 44°C IS 258ml .</u>

7 0
3 years ago
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