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Len [333]
2 years ago
12

Find the molarity of a 500 L solution that contains 10 moles of fluorine.

Chemistry
1 answer:
deff fn [24]2 years ago
3 0

<u>Given </u><u>:</u><u>-</u>

  • Number of moles = 10
  • Volume of solution = 500L

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

  • Molarity of the solution .

<u>Solution</u><u> </u><u>:</u><u>-</u>

As we know that ,

\longrightarrow Molarity (M )=\dfrac{Number\ of \ moles \ of \ solute }{Volume\ of \ solution\ (in \ L) }

Substitute ,

\longrightarrow M =\dfrac{10}{500L}

Simplify,

\longrightarrow M = \dfrac{1}{50}

Convert into decimal ,

\longrightarrow \underline{\underline{ M = 0.02 \ mol \ L^{-1}}}

This is the required answer .

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When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr
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<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

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                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

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K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

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