Calculate the molarity of the solute in a solution containing 14.2 KCl in 250 mL solution?

I know molarity is (moles of the solute/ liters of the solution)
The issue I dont understand is the 14.2 KCl I dont know what the units are. I am assuming its not grams

Answer 1

#"0.76 mol L"^(-1)#

I would say that you're dealing with a solution that contains #14.2# grams of potassium chloride, #"KCl"#, in #"250 mL"# of solution.

If so, your approach will be to determine the number of moles in the sample using the molar mass of potassium chloride.

#14.2 color(red)(cancel(color(black)("g"))) * "1 mole KCl"/(74.55color(red)(cancel(color(black)("g")))) = "0.1905 moles KCl"#

To put it simply, molarity is the number of moles of solute a solution contains per liter of solution, which indicates its concentration.

This means that in order to find a solution's molarity, you essentially must figure out how many moles of solute you have in #"1 L"# of solution.
In your case, you know that #"250 mL"#, which is equivalent to #1/4"th"# of a liter, contains #0.905# moles.
All you have to do now is scale up this solution so that its volume becomes #"1 L"#. Simply put, if #1/4"th"# of a liter contains #0.1905# moles, it follows that #"1 L"# will contain four times as many moles of solute.
#1 color(red)(cancel(color(black)("L solution"))) * "0.905 moles KCl"/(1/4color(red)(cancel(color(black)("L solution")))) = "0.76 moles KCl"#
So, if #"1 L"# of this solution contains #0.76# moles of solute, it follows that its molarity is
#"molarity" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.76 mol L"^(-1))color(white)(a/a)|)))#

The number of sig figs you have for the volume of the solution is the answer, which is rounded to two sig figs.

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Answer 2

To calculate the molarity (M) of the solute in a solution, you need to know the number of moles of the solute (in this case, KCl) and the volume of the solution in liters.

First, you need to convert the volume of the solution from milliliters (mL) to liters (L): [ 250 , \text{mL} = 250 \times 10^{-3} , \text{L} = 0.25 , \text{L} ]

Next, you need to calculate the number of moles of KCl: [ \text{moles of KCl} = \frac{\text{mass of KCl (in grams)}}{\text{molar mass of KCl}} ]

The molar mass of KCl is the sum of the atomic masses of potassium (K) and chlorine (Cl): [ \text{molar mass of KCl} = 39.10 , \text{g/mol (K)} + 35.45 , \text{g/mol (Cl)} = 74.55 , \text{g/mol} ]

Now, plug in the values to find the number of moles of KCl: [ \text{moles of KCl} = \frac{14.2 , \text{g}}{74.55 , \text{g/mol}} \approx 0.1907 , \text{moles} ]

Finally, calculate the molarity of the solution using the formula: [ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} ]

[ \text{Molarity (M)} = \frac{0.1907 , \text{moles}}{0.25 , \text{L}} = 0.7628 , \text{M} ]

Therefore, the molarity of the solute (KCl) in the solution is approximately 0.7628 M.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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