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2 years ago

5) Calculate the molality of 0.210 mol of KBr dissolved in 0.075kg pure water?

Chemistry

1 answer:

Margaret [11]2 years ago

5 0

Answer:

$\boxed {\boxed {\sf 2.8 \ m }}$

Explanation:

The formula for molality is:

$m=\frac{moles \ of \ solute}{kg \ of \ solvent}$

There are 0.210 moles of KBr and 0.075 kilograms of pure water.

$moles= 0.210 \ mol \\kilograms = 0.075 \ kg$

Substitute the values into the formula.

$m= \frac{ 0.210 \ mol }{0.075 \ kg}$

Divide.

$m= 2.8 \ mol/kg= 2.8 \ m$

The molality is <u>2.8 moles per kilogram</u>

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The kind of reaction that occurs when you mix aqueous solutions of barium sulfide and sulfuric acid is a precipitation reaction.

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Level: High school

Subject: Chemistry

Topic: Chemical reactions

Sub-topic: Precipitation reactions

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2 years ago

If you had a 0.5 M KCl solution, how much solute would you have in moles, and what would the solute be?

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Answer:

37.25 grams/L.

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Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.

<em>M = (no. of moles of KCl)/(volume of the solution (L))</em>

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∴ M = [(mass of KCl)/(molar mass of KCl)]/(volume of the solution (L))

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3 years ago

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2K + CrBr₂ → 2KBr + Cr

2mole 1 mole 1 mole

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From chemical equation

1 mole of CrBr₂ = 2 moles of K

∴ 0.939 moles of CrBr₂ = ?

⇒ 0.939 x 2/1 = 1.878 moles of K

1.878 moles of K is needed, but there is 1.458 moles of K. So, Potassium is completed first during the reaction . Hence, Potassium is limiting reagent. and CrBr₂ is excess reagent .

From chemical equation

2 moles of K = 1 mole of Cr

∴ 1.458 moles of K = ?

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3 years ago