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On mixing, heptane and octane form an id...

On mixing, heptane and octane form an ideal solution. At 373 K. The vapour pressure of the two luquid components (heptane and octane) are 105 kPa and 45 kPa repectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane= 100 g `mol^(-1)` and of octane= 114 g `mol^(-1)`)

A

144.5 kPa

B

72.0 kPa

C

36.1 kPa

D

96.2 kPa

Text Solution

Verified by Experts

The correct Answer is:
b
`n_("ethanol")=((25g))/((100mol^(-1)))=0.25 "mol"`
` X_("octane")=((35g))/((114 g mol^(-1)))=0.30 "mol"`
`X_("heptane")=(0.25)/(0.25+0.30)=0.45`
`"Xoctane"=(0.30)/(0,25+0.30)=0.55`
`P=(0.45xx105 "K pa")xx(0.55xx45"K pr")`
`=47.25+24.75=72 "K pa"`.
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