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Elanso [62]
3 years ago
15

A chemist needs to make 250 mL of a 2.50 M aqueous solution of ammonium hydroxide from a 6.00 M ammonium hydroxide solution. How

much water must be combined with a sample of the 6.00 M solution to make the diluted solution? The answer is 146 mL but I keep getting 104 mL through M1V1=M2V2. What must I additionally do?
Chemistry
1 answer:
Furkat [3]3 years ago
7 0

Answer:

250 mL (total solution)  = 104 mL (stock solution)  + 146 mL (water)

Explanation:

Data Given

M1 = 6.00 M

M2 = 2.5 M

V1 = 250 mL

V2 = ?

Solution:

As the chemist needs to prepare 250 mL of solution from 6.00 M ammonium hydroxide solution to prepare a 2.50 M aqueous solution of ammonium hydroxide.

Now

first he have to determine the amount of ammonium hydroxide solution that will be taken from6.00 M ammonium hydroxide solution

For this Purpose we use the following formula

                    M1V1=M2V2

Put values from given data in the formula

                   6 x V1 = 2.5 x 250

Rearrange the equation

                   V1 = 2.5 x 250 /6

                    V1 = 104 mL

So 104 mL is the volume of the solution which we have to take from the 6.00 M ammonium hydroxide solution to prepare 2.5 M  aqueous solution of ammonium hydroxide

But we have to prepare 250 mL of the solution.

so the chemist will take 104 mL from 6.00 M ammonium hydroxide solution and have to add 146 mL water to make 250 mL of new solution.

in this question you have to tell about the amount of water that is 146 mL

250 mL (total solution)  = 104 mL (stock solution)  + 146 mL (water)

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Correct Answer: option 1 i.e. C

Reason: 
The the compound of interest i.e.  XCl4, since there are 4 Cl atoms bonded to X. This signifies that the valency of X is 4.

There atomic number of C is 6. It's electronic configuration is giving by 1s2 2s2 2p2. Thus, there are 4 electrons in valence shell of C. This signifies that valency of C is 4. Hence the compound present in present case is CCl4.
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If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
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Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

<em>Consider the following reaction: </em>

<em>SO2Cl2 -----> SO2(g) + Cl2(g) </em>

<em>Kc= 2.99 x 10^-7 at 227 degrees celcius </em>

<em>If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?</em>

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

       SOCl2 ---------> SO2 + Cl2     Kc = 2.99x10⁻⁷

i)       0.168                  0        0

c)         -x                    +x       +x

e)     0.168-x                x         x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

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<em>[Cl₂] = 2.24x10⁻⁴ M</em>

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3 years ago
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What is molarity of 47.0 g KCl dissolved in enough water to give 375 mL of solution?
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This question provides us –

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__________________________________________

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\qquad \purple{\twoheadrightarrow\bf Molarity _{(Solution)} =  \dfrac{ W\times 1000}{MV}}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \dfrac{ 47 \times 1000}{74.5513\times 375}

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\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \cancel{\dfrac{47000}{27956.7375}}

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