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Part a cyclohexane has a freezing point of 6.50 ∘c and a kf of 20.0 ∘c/m. What is the freezing point of a solution made by disso

Andrei [34K]

The mathematical expression for the depression in freezing point is given as:

$\Delta T_{f}= k_{f}m$

where, $\Delta T_{f}$ = depression in freezing point

$k_{f}$ = molal depression constant (20.0 $^{o}Ckg/mole$ )

m = molality

Molality is defined as the ratio of number of moles of solute to the kg of the solvent

Number of moles = $\frac{given mass in g}{molar mass}$

given mass of biphenyl solute = 0.925 g

Molar mass of biphenyl = 154.21 g/mol

Put the values,

number of moles of biphenyl = $\frac{0.925 g}{154.21 g/mol}$

= $0.0059 mole$

Molality = $\frac{number of moles of biphenyl solute}{mass of the solvent in kg}$

Mass of cyclohexane solvent = 25.0 g

Convert g into kg: 1 kg = 1000 g

Thus, mass of cyclohexane = $\frac{25.0}{1000}$

= 0.025 kg

Now, put the values in the formula of molality:

Molality = $\frac{0.0059 mole}{0.025 kg}$

= 0.236 mole/kg or 0.236 m.

Calculate the depression in freezing point:

$\Delta T_{f}= 20.0^{o}C kg/mole\times 0.236 mole/kg$

= $4.72 ^{o}C$

Now, $\Delta T_{f} = T_{solvent} - T_{solution}$

$4.72 ^{o}C = 6.50^{o}C - T_{solution}$

$T_{solution} = 6.50^{o}C- 4.72 ^{o}C$

= $1.78^{o}C$

Hence, freezing point of the solution is $1.78^{o}C$ .

4 0

3 years ago

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Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago

Stearic acid, fatty acid commonly found in animal fat has a condensed Chemical formula CH3(CH2)16COOH. It is a common component

timofeeve [1]

Answer:

The correct answer is option C.

Explanation:

A. The -COOH group would experience hydrogen bonding with water.

The statement is true because , double bonded oxygen atom of carboxylic group interacts with hydrogen atom of water molecule and oxygen atom of water molecule is in same interaction with hydrogen atom of carboxylic acid.

B. The long alkane ''tail" $CH_3(CH_2)_{16} -$ , is hydrophobic.

The statement is true because , region which do not interacts with water molecule is termed as hydrophobic region and region which do interacts with water molecule is termed as hydrophilic region. Generally ,alkanes are are hydrophobic in nature so is the alkane chain in stearic acid.

C. Stearic acid and water are probably miscible.

The statement is false because of higher hydrophobic region the interaction between the acid molecules and water molecules results in immiscibility.

D. Stearic acid would probably dissolve in non polar solvents such as hexane $C_6H_{14}$ .

The statement is true because of hydrophobic part in stearic acid is larger so that is why it get easily dissolve in non polar solvents.

E. Stearic acid contains hydrophobic regions

The statement is true because of higher hydrophobic region due to which they easily get dissolved in organic and non polar solvents.

6 0

4 years ago

How long is breast milk good for out of the fridge?

Helen [10]

Answer:

until it gets bright white or smells bad, then throw it out

Explanation:

4 0

3 years ago

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Nitric acid is usually purchased in a concentrated form that is 70.3% HNO3 by mass and has a density of 1.41 g/mL. How much conc

Marina CMI [18]

Answer:

The answer to your question is: 9 .43 ml of HNO3 70.3%

Explanation:

Data

Purity: 70.3%

density = 1.41g/ml

Volume = 1 l

Concentration = 0.12 M

Molecular weight = 1 + 14 + (16 x 3) = 63 gr

63 gr of HNO3 ------------------1 mol of HNO3 or IM

x ------------------- 0.12 M

x = 0.12M x 63 /1 = 7.65 g of HNO3

Now, calculate volume

density = mass / volume and volume = mass/density

volume = 7.65 / 1.41 = 6.63 ml of HNO3

Now, consider purity (it's and inver rule of three)

6.63 ml ----------------- 100%

x ----------------- 70.3 %

x = (6,63 x 100)/70.3 = 9.43 ml of HNO3

8 0

4 years ago