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If Delta H("vaporisation") of (C(2)H(5))...

If `Delta H_("vaporisation")` of `(C_(2)H_(5))O(l)` is `350 J//g` at it's boiling point 300 K, then molar entropy change for condensation process is

A

`86.33 J//mol.K`

B

`-86.33J//mol.K`

C

`-1.16 J//mol.K`

D

`1.16`

Text Solution

Verified by Experts

The correct Answer is:
B

`DeltaH_("vaporisation")=350xx74" J/mol"`
`DeltaS_("condensation")=(-DeltaH_("vaporisation"))/(T_("boiling point"))`
`=(-350xx74)/(300)" J/mol.K "=-86.33" J/mol.K"`
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Assertion (A): The enthalpy of formation of H_(2)O(l) is greater than that of H_(2)O(g) . Reason (R ) : Enthalpy change is negative for the condensation reaction H_(2)O(g) rarr H_(2)O(l) .

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