What is the boiling point of an aqueous solution of $L i B r$ $LiBr$ that is $0.527 \cdot m o l \cdot k g^{- 1}$ $0.527molkg^-1$ with respect to the salt?

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What is the boiling point of an aqueous solution of $L i B r$ $LiBr$ that is $0.527 \cdot m o l \cdot k g^{- 1}$ $0.527molkg^-1$ with respect to the salt?

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Answer 1 · 2017-08-09T18:53:24

This site lists $K_{B}$ $K_B$ , the ebullioscopic constant of water, as $0.512 \cdot^{\circ} C \cdot k g \cdot m o l^{- 1}$ $0.512*""^@C*kg*mol^-1$ .....

Explanation:

We know that in aqueous solution the lithium bromide salt speciates to its constituent ions.........

$L i B r (s) H_{2} O - - \to L i^{+} + B r^{-}$ $LiBr(s) stackrel(H_2O)rarrLi^+ + Br^(-)$

And $DeltaT_"boiling point elevation"=ixx"molal conc."xxK_B$

And $i=2$ because of the speciation of the ionic solute......

$DeltaT=2xx0.527*mol*kg^-1xx0.512*kg*mol^-1*""^@C$

$=0.540$ $""^@C$

And thus the observed boiling point of the solution should be $100.540$ $""^@C$ .

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What is the boiling point of an aqueous solution of $L i B r$ $LiBr$ that is $0.527 \cdot m o l \cdot k g^{- 1}$ $0.527molkg^-1$ with respect to the salt?

1 Answer

Explanation:

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What is the boiling point of an aqueous solution of LiBrLiBrLiBr that is 0.527*mol*kg^-10.527⋅mol⋅kg−10.527*mol*kg^-1 with respect to the salt?

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Explanation:

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What is the boiling point of an aqueous solution of $L i B r$ $LiBr$ that is $0.527 \cdot m o l \cdot k g^{- 1}$ $0.527molkg^-1$ with respect to the salt?