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The vapour pressure of pure benzene at c...

The vapour pressure of pure benzene at certain temperature is 0.850 bars. A non-volatile, non-electrolyte solid weighing 0.5 grams when added to 39 grams of benzene (molar mass 78grams), vapour pressure of the solution becomes 0.845 bars. What is the molar mass of the solid substance? 

Text Solution

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Vapoure pressure of solvent benzene=`P^(0)=0.850` bar
Vapour pressure off solution=P=0.845 bar.
Lowering of vapour pressure=`P^(0)-P=0.850-0.845=0.005` bar
`(P^(0)-P)/(P^(0))=(0.005)/(0.850)=0.00588`
Mass of solute=`W_(2)=0.5g, " Mass of solvent benzene"W_(1)=39g`
Molar mass of benzene=78g `mol^(-1)`
Formula: `M_(2)=(W_(2))/(W_(1))xx(M_(1))/([(P^(0)-P)/(P^(0))])`
Substitution `M_(2)=(0.5)/(39)xx(78)/(0.0059)`
Answer `M_(2)=170g//mol`.
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The vapour pressure of pure benzene at a certain tempreature is 0.850 bar. A non-volatile. non-electrolyte solid weighing 0.5g, when added to 39.0g of benzene (molar mass of benzene 78g mol^-1 ) vapour pressure of the solution then is 0.845 bar. What is the molar mass of the solid substance?

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Knowledge Check

  • 1.5 g of a non-volatile, non-electrolyte is dissolved in 50 g benzene ( K_b = 2.5 kg mol^(-1) ). The elevation of the boiling point of the solution is 0.75 K. The molecular weight of the solute in g mol^(-1) is :

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    B
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