Since we were only given the (M) and grams that LiBr we had actually tofind the mol the solute. To discover the #mol that solute we split thegrams (which to be given) end the molar massive of LiBr (87). Us thengot 1.15 . Now we room able to discover the Liter of systems . An initial weplugged in our offered values M= 4 & #mol solute= 1.15 . Come findliter of systems we just dived 1.15 end 4 which equals0.28.


I expect that lithium bromide is no so dissolve to prepare 4 Msolutions in water at 20 0C.

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∙ 10y ago
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