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kogti [31]
3 years ago
5

A) How do you prepare %3 (w/v) Na2CO3 solution from Na2CO3⸱2H2O? (15p) Na2CO3 MW=106 g/mol

Chemistry
1 answer:
just olya [345]3 years ago
8 0

Answer:

4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL

Explanation:

A 3%(w/v) solution contains 3g of solute (In this case, Na2CO3) in 100mL of solution.

Assuming we require 100mL of solution we must add 3g of Na2CO3. The reactant that is available is its dihydrate, with molar mass:

106g/mol + 2*MW H2O

106g/mol + 2*18g/mol = 142g/mol

That means the mass of Na2CO3.2H2O that must be added to prepare the solution is:

3g Na2CO3 * (142g/mol Na2CO3.2H2O / 106g/mol Na2CO3) =

<h3>4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL</h3>
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Explanation : Given,

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or,

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Now put all the given values in this formula, we get:

0.349=\frac{4\times 55.85g/mol}{\text{molar mass of hemoglobin}}\times 100

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Thus, the molar mass of hemoglobin is, 6.40\times 10^4g/mol

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A magnesium ion, Mg2+, is formed from a magnesium atom, Mg.
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Answer:

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I hope it helps you!                            

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<h3>What is Titration ? </h3>

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