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2 years ago

What is the molarity of a solution containing 3.6 mol of KCl in 750 mL?

Chemistry

1 answer:

Vesnalui [34]2 years ago

4 0

Answer:

$\boxed {\boxed {\sf 4.8 \ M \ KCl}}$

Explanation:

The molarity of a solution is found by dividing the moles of solute by the liters. of solvent.

$molarity=\frac{moles}{liters}$

We know the solution has 3.6 moles of potassium chlorine. We know there are 750 milliliters of solvent, but we need to convert this to liters.

1 liter is equal to 1000 milliliters. Set up a proportion.

$\frac {1 \ L }{1000 \ mL}$

Multiply by 750 mL and the units of mL will cancel.

$750 \ mL * \frac {1 \ L }{1000 \ mL} = \frac {750 \ mL}{1000 \ mL}= 0.75 \ L$

Now we know the moles and liters, so we can calculate molarity.

moles= 3.6 mol KCl
liters= 0.75 L

Substitute these values into the formula.

$molarity= \frac{3.6 \ mol \ KCl}{0.75 \ L}$

Divide.

$molarity= 4.8 \ mol \ KCl/ L$

1 mole per liter is equal to 1 molar.
Our answer of 4.8 moles of potassium chloride per liter is equal to 4.8 M KCl

$molarity= 4.8 \ M \ KCl$

The molarity of the solution is 4.8 M KCl.

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Answer:

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General Formulas and Concepts:

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Explanation:

Step 1: Define

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2 years ago

The measured voltage of an electrochemical cell consisting of pure nickel immersed in a solution of Ni2+ ions of unknown concent

Verizon [17]

Answer:

[Ni²⁺] = 1.33 M

Explanation:

To do this, we need to use the Nernst equation which (in standard conditions of temperature)

E = E° - RT/nF lnQ

However R and F are constant, and the reaction is taking place in 25 °C so we can assume the nernst equation like this:

E = E° - 0.05916/n logQ

As the nickel is in the cathode, this means that this element is being reducted while Cadmium is being oxidized, therefore the REDOX reaction would be:

Cd(s) + Ni²⁺(aq) --------> Cd²⁺(aq) + Ni(s)

With this, Q:

Q = [Cd²⁺] / [Ni²⁺]

Now, we need to know the value of the standard reduction potentials, which can be calculated with the semi equations of reduction and oxidation:

Cd(s) ------------> Cd²⁺ + 2e⁻ E°₁ = 0.40 V

Ni²⁺ + 2e⁻ -------------> Ni(s) E°₂ = -0.25 V

E° = E°₁ + E°₂

E° = 0.40 - 0.25 = 0.15 V

Now that we have all the data, we can solve for the [Ni²⁺]:

0.133 = 0.15 - 0.05916/2 log(5/[Ni²⁺])

0.133 - 0.15 = -0.05916/2 log(5/[Ni²⁺])

-0.017 = -0.02958 log(5/[Ni²⁺])