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3 years ago

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Hexane and octane are mixed to form a 45 mol% hexane solution at 25 deg C. The densities of hexane and octane are 0.655 g/cm3 an

d 0.703 g/cm3, respectively. Assume you have 1.0 L of octane. Calculate the required volume of hexane. Report your answer in liters.

1 answer:

avanturin [10]3 years ago

5 0

Answer:

The required volume of hexane is 0.66245 Liters.

Explanation:

Volume of octane = v=1.0 L= $1000 cm^3$

Density of octane= d = $0.703 g/cm^3$

Mass of octane ,m= $d\times v=0.703 g/cm^3\times 1000 cm^3=703 g$

Moles of octane = $\frac{m}{114 g/mol}=\frac{703 g}{114 g/mol}=6.166 mol$

Mole percentage of Hexane = 45%

Mole percentage of octane = 100% - 45% = 55%

$55\%=\frac{6.166 mol}{\text{Total moles}}\times 100$

Total moles = 11.212 mol

Moles of hexane :

$45%=\frac{\text{moles of hexane }}{\text{Total moles}}\times 100$

Moles of hexane = 5.0454 mol

Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g

Density of the hexane,D = $0.655 g/cm^3$

Volume of hexane = V

$V=\frac{M}{D}=\frac{433.9044 g}{0.655 g/cm^3}=662.4494 cm^3\approx 0.66245 L$

(1 cm^3= 0.001 L)

The required volume of hexane is 0.66245 Liters.

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Answer:

Stock solution = 1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

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Explanation:

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

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

<u>Step 2</u>: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

<u>Step 3:</u> Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

<u>Step 4</u>: Calculate dilution 2

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M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

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M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

<u>Step 6</u>: Calculate dilution 4

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For the given equation:

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Putting values in above equation, we get:

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We are given:

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Putting values in above equation, we get:

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$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

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