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kherson [118]
3 years ago
15

What is the volume of the solution produced when enough water is added to 42.0 g of mgcl2 ⋅ 6h2o to yield a solution that has a

cl- ion concentration of 0.500m?
Chemistry
1 answer:
DENIUS [597]3 years ago
5 0

Answer:

827 mL

Explanation:

To answer this question we use the <em>definition of Molarity</em>:

Molarity = mol / L

[Cl⁻] = mol Cl⁻ / L

Now we calculate the moles of Cl⁻ present in 42.0 g of MgCl₂⋅6H₂O:

Molar mass of MgCl₂⋅6H₂O = 24.3 + 2*35.45 + 6*18 = 203.2 g/mol

moles of Cl⁻ = 42.0 g MgCl₂⋅6H₂O ÷ 203.2 g/mol * \frac{2molCl^{-}}{1molMgCl_{2}.6H_{2}O} = 0.4134 mol Cl⁻

Finally we use the definition of molarity to calculate the volume:

0.500 M = 0.4134 mol Cl⁻ / xL

xL = 0.827 L =  827 mL

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Consider the generic reaction: 2 A(g) + B(g) → 2 C(g). If a flask initially contains 1.0 atm of A and 1.0 atm of B, what is the
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An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?
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<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

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\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

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\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

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m_{solute} = Given mass of solute (CsCl) = 8.00 g

M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

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\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

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