Home
Class 12
CHEMISTRY
The molal freezing point depression cons...

The molal freezing point depression constant of benzene`(C_(6)H_(6))` is `4.90 K kg mol^(-1)`. Selenium exists as a polymer of the type `Se_(x)`. When `3.26 g` of selenium is dissolved in `226 g` of benzene, the observed freezing point is `0.112^(@)C` lower than pure benzene. Deduce the molecular formula of selenium. (Atomic mass of `Se=78.8 g mol^(-1)`)

Text Solution

Verified by Experts

Calculation of milar mass of selenimum `(Se_(x))`
`M_(B)-(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))`
Mass of selenium `(W_(B))=3.26g`
Mass of benzene`(W_(A))226 g=0.226 kg`
Depression in freezing point `(DeltaT_(f))=0.112^(@)C=0.112 K`
Molal depression constant `(K_(f))=4.9 K kg mol^(-1)`
`M_(B)=((4.9K kg mol^(-1))xx(3.26g))/((0.112K)xx(0.226 kg))=631.08 g mol^(-1)`
Gram atomic of selenium = `78.8 g mol^(-1)`
In `Se_(x)x(78.8 g mol^(-1))=631.08 g mol^(-1)`
`x=((631.08 g mol^(-1)))/((78.8g mol^(-1)))=8`
Molecular fromula f selenum=`Se_(8)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

The molal freezing point depression constant for benzene is 4.9 K kg mol^(-1) . Selenium exists as a polymer of the type Se_(n) . When 3.26 g selenium is dissolved in 226 g of benzene, the observed freezing. If molecular formula of sulphur is S_(n) . Then find the value of n . (at. wt. of S = 32 ).

0.440 g of a substance dissolved in 22.2 g of benzene lowered the freezing point of benzene by 0.567 ^(@)C . The molecular mass of the substance is _____ (K_(f) = 5.12 K kg mol ^(-1))

The freezing point depression constant of benzene is 5.12 K kg mol^(-1) . The freezing point depression for the solution of molality 0.078m containing a non-electrolyte solute in benzene is

On adding 1g arsenic to 80 g benzene, the freezing point of benzene is lowered by 0.213^(@)C . The formula of arsenic molecule is : (Atomic mass of arsenic = 75) ( K_(f) = 5.12 K kg/mol)