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alexandr1967 [171]
3 years ago
14

I have it attached plz help ASAP

Chemistry
1 answer:
givi [52]3 years ago
4 0

Answer:

The answer would most likely be C

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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
avanturin [10]

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.

5 0
3 years ago
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