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4 years ago

Determine the freezing point of an aqueous solution containing 10.50 g of magnesium bromide in 200.0 g of water.

1 answer:

Rudiy274 years ago

5 0

For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:

ΔTf = (Kf)(m)(i)
where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3

Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol

m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg

For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf

Tf = -1.59 celsius

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3 years ago

[20pts] I want to start counterfeiting American pennies. Each penny weighs 2.56g and I want $200.00 of pennies for my scheme. Se

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Answer:

$\$200.00\times 100\dfrac{penny}{\$}\times \dfrac{2.56g}{penny}\times \dfrac{1molCu}{63.5gCu}\times \dfrac{2molAl}{3molCu}\times \dfrac{27gAl}{1molAl}$

Explanation:

The dimensional analysis helps you by focusing your attention into the units: you must set up the equation starting with the available units and add the factors with the units such that the cancelation of numerators and denominators ends up with the desired units: grams of aluminium in this case.

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You must also use the atomic masses of Cu and Al which are 63.5g/mol and 27g/mol respectively.

The equation is:

$\$200.00\times 100\dfrac{penny}{\$}\times \dfrac{2.56g}{penny}\times \dfrac{1molCu}{63.5gCu}\times \dfrac{2molAl}{3molCu}\times \dfrac{27gAl}{1molAl}$

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3 years ago

What is the molarity of a solution prepared by diluting 250 mL of a 40% H2SO4 solution to 1 Liter? The density of the stock solu

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Answer:

1.195 M.

Explanation:

We can calculate the concentration of the stock solution using the relation:

M = (10Pd)/(molar mass).

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d is the density of H₂SO₄ (d = 1.17 g/mL).

molar mass of H₂SO₄ = 98 g/mol.

∴ M of stock H₂SO₄ = (10Pd)/(molar mass) = (10)(40%)(1.17 g/mL) / (98 g/mol) = 4.78 M.

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∴ (MV) before dilution = (MV) after dilution

M before dilution = 4.78 M, V before dilution = 250 mL.

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∴ M after dilution = (MV) before dilution/(V after dilution) = (4.78 M)(250 mL)/(1000 mL) = 1.195 M.

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3 years ago

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