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3 years ago

What is the molarity of a sodium chloride solution made by dissolving 4.512 moles to make 2.0 L?

Chemistry

1 answer:

Svet_ta [14]3 years ago

6 0

Answer:

2.40 M

Explanation:

The molarity of a solution tells you how many moles of solute you get per liter of solution.

Notice that the problem provides you with the volume of the solution expressed in milliliters,

. Right from the start, you should remember that you must convert this volume to liters by using the conversion factor

1 L

Now, in order to get the number of moles of solute, you must use its molar mass. Now, molar masses are listed in grams per mol,

g mol

−

, which means that you're going to have to convert the mass of the sample from milligrams to grams

1 g

Sodium chloride,

NaCl

, has a molar mass of

58.44 g mol

−

, which means that your sample will contain

unit conversion



280.0

⋅

molar mass



1 mole NaCl

58.44

0.004791 moles NaCl

This means that the molarity of the solution will be

solute

solution

0.004791 moles

2.00

⋅

−

2.40 M

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

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What is the total number of atoms in 0.20 mol of propanone, CH3COCH3?

Nutka1998 [239]

Answer:

1.2×10²³ atoms.

Explanation:

Data obtained from the question include:

Mole of propanone = 0.20 mole

Number of atoms of propanone =.?

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ atoms.

This implies that 1 mole of propanone also contains 6.022×10²³ atoms.

Thus, we can obtain the number of atoms in 0.20 mole of propanone as illustrated below:

1 mole of propanone contains 6.022×10²³ atoms.

Therefore, 0.20 mole of propanone will contain = 0.2 × 6.022×10²³ = 1.2×10²³ atoms.

Thus, 0.20 mole of propanone contain

1.2×10²³ atoms.

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3 years ago

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enyata [817]

Answer:

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3 years ago

If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.

kicyunya [14]

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$CH_{3}COOH \ \textless \ ---\ \textgreater \ H^{+} + CH_{3}COO^{-}$

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

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The change in Concentration Δ $[CH_{3}COOH]$ = 0.001 $moldm^{-3}$

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Change $moldm^{-3}$ -0.001 +0.001 +0.001

Equilibrium $moldm^{-3}$ $x$ - 0.001 0.001 0.001


Since the $k_{a}$ value is so small, the assumption
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$k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}$

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps!

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