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Angelina_Jolie [31]

2 years ago

10

A solution is prepared by dissolving 215 grams of methanol, ch3oh, in 1000. grams of water. what is the freezing point of this s

olution? [the freezing point depression constant for water is 1.86°c/mole solute in 1000g of water]

1 answer:

Leya [2.2K]2 years ago

6 0

Answer : The freezing point of the solution is, 260.503 K

Solution : Given,

Mass of methanol (solute) = 215 g

Mass of water (solvent) = 1000 g = 1 kg (1 kg = 1000 g)

Freezing depression constant = $1.86^oC/m=1.86Kkg/mole$

Formula used :

$\Delta T_f=K_f\times m\\T^o_f-T_f=K_f\times \frac{w_{solute}}{M_{solute}\times w_{solvent}}$

where,

$T^o_f$ = freezing point of water = $100^oC=273K$

$T_f$ = freezing point of solution

$K_f$ = freezing point constant

$w_{solute}$ = mass of solute

$w_{solvent}$ = mass of solvent

$M_{solute}$ = molar mass of solute

Now put all the given values in the above formula, we get

$273K-T_f=(1.86Kkg/mole)\times \frac{215g}{(32g/mole)\times (1kg)}$

By rearranging the terms, we get the freezing point of solution.

$T_f=260.503K$

Therefore, the freezing point of the solution is, 260.503 K

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2 years ago

Calculate the activity coefficients for the following conditions:

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<u>Answer:</u>

<u>For a:</u> The activity coefficient of copper ions is 0.676

<u>For b:</u> The activity coefficient of potassium ions is 0.851

<u>For c:</u> The activity coefficient of potassium ions is 0.794

<u>Explanation:</u>

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

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where,

$C_i$ = concentration of i-th ions

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<u>For a:</u>

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

$C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M$

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$Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{ (known)}\\\mu=0.01M$

Putting values in equation 1, we get:

$-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676$

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<u>For b:</u>

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

$C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.025M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851$

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<u>For c:</u>

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Putting values in equation 2, we get:

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Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.06M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794$

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Answer:

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Explanation:

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b)

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