1.

The vapour pressure of CS_(2) at 50^(@)C is 845 torr and a solution of 2.0 g sulphur in 100g of CS_(2) has a vapour pressure of 845/9 torr. Calculate the formula of sullphur molecule.

Answer»

Solution :Calculation of molar mas of sulphur.
Accoring to RAOULT's law, `(P_(A)^(@)-P_(S))/P_(S)=(W_(B)xxM_(A))/(M_(B)xxW_(A))`
`P_(A)^(@)854 TORR, P_(S)848.9 torr, W_(B)=2.0g,`
`W_(A)=100 g, M_(A)(CS_(2))=76 g mol^(-1)`
`((854torr-848.9torr))/((848.9torr))=((2.0g)xx(76.g mol^(-1)))/(M_(B)xx(100g))`
`M_(B)=((2.0g)xx(76.g mol^(-1))xx(848.9 torr))/((5.1 torr)xx(100g))=253.0 mol^(-1)`
Calculation of molecular formula of sulphur.
LET the formula of sulphur=`S_(N)`
`NXX(32g mol^(-1))=253.0 g mol^(-1)`
`n=((253g mol^(-1)))/((32g mol^(-1)))=7.9~~8`
Thus, the molecular formula of suphur=`S_(8)`.


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