What is the mole fraction of $LiCI$ in a mixture of 0.564g $NaCl$ , 1.52g $KCl$ and 0.857g $LiCI$ ?

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Answer 1 · 2017-07-11T01:57:46

Mole fraction $(X_(compound))$ = $("moles compound"/"total moles of all compounds")$

Given:
$0.564 g NaCl$ = $(0.564g)/(58(g/"mol"))$ = $0.0097"mole"$

$1.52 g KCl$ = $(1.52g)/(74(g/"mol"))$ = $0.0205"mole"$

$0.857 g LiCl$ = $(0.857g)/(42(g/"mol"))$ = $0.0204"mole"$

$Sigma "moles" = (0.0097 + 0.0205 + 0.0204)"mole"$ = $"0.0506mole"$

$X_(LiCl)$ = $("moles LiCl")/(Sigma"moles")$ = $(0.0204)/(0.0506)$ = $0.04032$

What is the mole fraction of LiCILiCI in a mixture of 0.564g NaClNaCl, 1.52g KClKCl and 0.857g LiCILiCI?