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Zina [86]
4 years ago
10

1718 l of a 0.3556-m c3h7oh solution is diluted to a concentration of 0.1222 m, what is the volume of the resulting solution

Chemistry
1 answer:
Agata [3.3K]4 years ago
5 0
Answer : The volume of the resulting solution will be 0.4999 L.

Explanation : As the two molar concentration of the C_{3} H_{7}OH solution are given and one of the volume concentration needs to be found.

So, according to the formula :-  m_{1}V_{1} = m_{2}V_{2}

And here if we consider V_{1} as 0.1718 L moles and m_{1}  as 0.3556 moles then we need to find V_{2} as m_{2}  is given as 0.1222 moles.
Hence on solving the formula we get,  

 V_{2} = (0.1718 X 0.3556) / 0.1222 = 0.4999 L

 


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Suppose a 500.mL flask is filled with 1.9mol of NO3 and 1.6mol of NO. The following reaction becomes possible: NO3gNOg 2NO2g The
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There is an error in the first sentence of the  question; the right format is:

Suppose a 500.mL flask is filled with 1.9mol of NO3 and 1.6mol of NO2.

It should be NO2 and not NO.

Answer:

The equilibrium molarity of NO = 0.21695 m

Explanation:

Given that :

the volume = 500 mL = 0.500 m

number of moles of NO_3 = 1.9 \ mol

number of moles of NO_2 = 1.6 \ mol

Then we can calculate for their respectively concentrations as :

[NO_3] = \frac{number \ of \ moles}{volume}

[NO_3] = \frac{1.9}{0.500}

[NO_3] = 3.8 \ M

[NO_2] = \frac{number \ of \ moles}{volume}

[NO_2] = \frac{}{} \frac{1.6}{0.500}

[NO_2] = 3.2 \ M

The chemical reaction can be written as:

NO_3_{(g)} + NO_{(g)} \to 2NO_2_{(g)}

The ICE table is as follows;

                    NO_3_{(g)} + NO_{(g)} \to 2NO_2_{(g)}

Initial              3.8         -               3.2

Change          +x          x               -2x

Equilibrium     3.8+x    +x              3.2 - 2x

K_c=\frac{[NO_2]^2}{[NO_3][NO]}

where \ K_c = 8.33

8.33 = \frac{(3.2-2x)^2}{(3.8+2x)x} \\ \\ 8.33 = \frac{(3.2-2x)^2}{(3.8x+2x^2)}

8.33(3.8x + 2x^2) = (3.2-2x)^2 \\ \\ 31.654x + 16.66x^2 = (3.2-2x)(3.2-2x) \\ \\ 31.654x + 16.66x^2 = 10.24 - 12.8x +4x^2 \\ \\ 10.24 - 44.454x -12.66 x^2 = 0 \\ \\ 12.66x^2 +44.454x -10.24 = 0

Using quadratic formula;

\frac{-b\pm \sqrt{(b)^2-4ac} }{2a}

= \frac{-(44.454) + \sqrt{(44.454)^2-4(12.66)(-10.24)} }{2(12.66)} \ \ OR \ \ \frac{-(44.454) - \sqrt{(44.454)^2-4(12.66)(-10.24)} }{2(12.66)}

= 0.21695 OR -3.7283

Going by the positive value;

x = 0.21695

[NO_3] = 3.8 +x  = 3.8 + 0.21695

= 4.01695 m

[NO] = x  = 0.21695 m

[NO_2] = 3.2 +x  = 3.2 + 0.21695

= 3.41695 m

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Answer:

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