Heptane and Octane form ideal solution at 373 K ,the vapour pressure of the two liquid components are 105.2 KPa and 46.8 kPa respectively what will be the vapour pressure of a mixture of 26.0 gram of 15 and 35 gram of Octane??
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moles of heptane = 26.0 /100
= 0.26
moles of Octane= 35.0/144 = 0.31
x(heptane) = 0.26 / 0.26 + 0.31
= 0.456
x(octane) = 0.31/0.26+0.31
= 0.544
total pressure= 105.2 × 0.456 + 46.8 × 0.544
= 47.97 + 25.46
= 73.43 kPa
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= 0.31
x(heptane) = 0.456
x(octane) = 0.544
total pressure= 105.2 × 0.456 + 46.8 × 0.544
= 47.97 + 25.46
= 73.43 kPa
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