How many grams of solute are present in 935 mL of 0.720 M $KBr$ ?

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Answer 1 · 2016-07-10T16:46:09

$"Grams of KCl"$ $~=$ $50*g$

$"Concentration"="Moles of solute"/"Volume of solution"$ . This gives units of $mol*L^-1$ .

Thus $"moles of solute"="volume"xx"concentration".=$

$935*cancel(mL)xx10^-3cancelL*cancel(mL^-1)xx0.729*cancel(mol)*cancel(L^-1)xx74.55*g*cancel(mol^-1)$

$=$ $??g$

How many grams of solute are present in 935 mL of 0.720 M KBrKBr?