What molar quantity of potassium bromide is present in a $68.5mL$ volume of a solution whose concentration is $1.65mol*L^-1$ with respect to the salt?

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What molar quantity of potassium bromide is present in a $68.5mL$ volume of a solution whose concentration is $1.65mol*L^-1$ with respect to the salt?

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Answer 1 · 2016-08-11T17:55:27

Approx. $0.1$ $mol$ $KBr$

Explanation:

$"Molarity"$ $=$ $"Moles of solute"/"Volume of solution"$ .

Thus $"moles of KBr"$ $=$ $68.5xx10^-3cancelLxx1.65*mol*cancel(L^-1)$ $=$ $0.113*mol$ .

I assume (perhaps wrongly!) that you quoted a $"molar"$ concentration (i.e. $mol*L^-1$ ) with respect to $KBr$ . You might have specified a $"molal concentration"$ , i.e. $"moles of solute"/"kilogram of solvent"$ , which is what " $"m"$ " generally signifies. The answers would be similar.

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What molar quantity of potassium bromide is present in a $68.5mL$ volume of a solution whose concentration is $1.65mol*L^-1$ with respect to the salt?

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What molar quantity of potassium bromide is present in a 68.5*mL68.5*mL volume of a solution whose concentration is 1.65*mol*L^-11.65*mol*L^-1 with respect to the salt?

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What molar quantity of potassium bromide is present in a $68.5mL$ volume of a solution whose concentration is $1.65mol*L^-1$ with respect to the salt?