1.

At 50^(@)C, the vapour pressure of pure CS_(2) is 854 torr. A solution of 2.0 g of sulphur in 100 gof CS_(2) has vapour pressure of 848.9 torr. Determine the formula of sulphur molecule.

Answer»


Solution :`(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1)) ""therefore""(854-848.9)/(854)=(2//M_(2))/(100//76)""("Molecular mass of CS"_(2)=12+2xx32=76u)`
`"or"(5.1)/(854)=(2)/(M_(2))XX(76)/(100)"or"M_(2)=254.5u`
If formula of SULPHUR molecules is `S_(x)`, molecular mass `=x xx 32.` Hence, `32x=254.5 or x = 8.`


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