A solution of `2.8g` of `Cdl_(2)` molar mass `=364g "mol"^(-1)`) in `20g` water has elevation in boiling point of `0.20^(0)`. What is molecular state

Question

A solution of `2.8g` of `Cdl_(2)` molar mass `=364g "mol"^(-1)`) in `20g` water has elevation in boiling point of `0.20^(0)`. What is molecular state

A solution of `2.8g` of `Cdl_(2)` molar mass `=364g "mol"^(-1)`) in `20g` water has elevation in boiling point of `0.20^(0)`. What is molecular state of `Cdl_(2)` in aqueous solution ? `[K_(b)(H_(2)O)=0.52^(0) "mol"^(-1)kg]`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - To decide molecular state, determine value of `i` or its molar mass.
(A) If `i=1`, and molar mass `=` calculated molar mass then solute remains without change
(B) If `igt1`, and molar mass `gt` caluclated molar mass then solute is ionised.
(C) If `ilt1`, and molar mass `gt` calculated molar mass then solute is polymerised (dimer, trimer, etc. )
`DeltaT_(b)(1000K_(b)w_(i))/(m_(1)w_(2))(i)`
`i=(DeltaT_(b)m_(1)w_(2))/(1000K_(b)w_(1))=(0.20xx364xx20)/(1000xx0.52xx208)=1.0`
Thus, `Cdl_(2)` remains `Cdl_(2)` (without ionisation) in aqueous solution.