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2 years ago

Suppose you have a sample of table salt (NaCl) that has a mass of 100.0 grams. If

Chemistry

1 answer:

IgorC [24]2 years ago

5 0

Taking into account the definition of molarity, the concentration of the solution is 0.855 $\frac{moles}{liter}$ .

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$Molarity=\frac{number of moles}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

<h3>Molarity of NaCl</h3>

In this case, you have:

number of moles of NaCl= $100 grams\frac{1 mole}{58.45 grams} =$ 1.71 moles (being 58.45 g/mole the molar mass of NaCl)
volume 2 L

Replacing in the definition of molarity:

$Molarity=\frac{1.71 moles}{2 L}$

Solving:

Molarity= 0.855 $\frac{moles}{liter}$

Finally, the concentration of the solution is 0.855 $\frac{moles}{liter}$ .

Learn more about molarity:

<u>brainly.com/question/9324116</u>

<u>brainly.com/question/10608366</u>

<u>brainly.com/question/7429224</u>

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Consider the electrolysis of molten barium chloride, BaCl2. (a) Write the half-reactions. (b) How many grams of barium metal can

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Answer: a) Cathode(-): $Ba^+^2+2e^-\rightarrow Ba$

Anode(+): $2Cl^-\rightarrow Cl_2+2e^-$

b) 0.640 grams of Ba will be deposited.

Explanation: a) The problem is based on Faraday law of electrolysis. Molten barium chloride has $Ba^+^2$ ion and $Cl^-$ ion. Barium ion is reduced and chloride is oxidized. Reduction takes place at cathode and oxidation at anode. So, the half reactions will be:

Cathode(-): $Ba^+^2+2e^-\rightarrow Ba$

Anode(+): $2Cl^-\rightarrow Cl_2+2e^-$

b) The question asks, how many grams of barium metal can be produced by supplying 0.50 ampere for 30 minutes.

From the Cathode half-reaction, 1 mole of Ba is deposited by 2 moles of electrons and we know that 1 mole of electron carries one Faraday that is 96485 Coulomb.

Coulombs for 2 moles of electrons will be = 2*96485 C = 192970 C

So, we can say that, 192970 C will deposit 1 mole of Ba metal.

Total available coulombs can be calculated using the formula:

q=i*t

where, q is electric charge in coulomb, i is current in ampere and t is time in seconds.

$q=q=0.50A*30min(\frac{60sec}{1min})$

q = 900 C (note: 1 C = 1 A*sec)

Let's calculate how many moles of Ba will get deposited by 900 C.

$900C(\frac{1molBa}{192970C})$

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Convert the moles of Ba to grams and for this we multiply by molar mass of Ba which is 137.33 gram per mol.

$0.00466molBa(\frac{137.33g}{1mol})$

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