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3 years ago

A solution is prepared by condensing 4.00 L of a gas,

Chemistry

1 answer:

vredina [299]3 years ago

5 0

Answer:

-2.3 ºC

Explanation:

Kf (benzene) = 5.12 ° C kg mol – 1

1st - We calculate the moles of condensed gas using the ideal gas equation:

n = PV / (RT)

P = 748/760 = 0.984 atm

T = 270 + 273.15 = 543.15 K

V = 4 L

R = 0.082 atm.L / mol.K

n = (0.984atm * 4L) / (0.082atm.L / K.mol * 543.15K) = 0.088 mol

Then, you calculate the molality of the solution:

m = n / kg solvent

m = 0.088 mol / 0.058 kg = 1.52mol / kg

Then you calculate the decrease in freezing point (DT)

DT = m * Kf

DT = 1.52 * 5.12 = 7.8 ° C

Knowing that the freezing point of pure benzene is 5.5 ºC, we calculate the freezing point of the solution:

T = 5.5 - 7.8 = -2.3 ºC

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Answer:

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Explanation:

<u>Step 1:</u> Data given

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3 years ago

A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at

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Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $heat_{absorbed}=heat_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 45 g

$m_2$ = mass of coffee = 180 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $85^oC$

$c_1$ = specific heat of aluminium = $0.80J/g^oC$

$c_2$ = specific heat of coffee= $4.186 J/g^oC$

Putting all the values in equation 1, we get:

$45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]$

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So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

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3 years ago

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Answer:

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