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12345 [234]
2 years ago
3

describe how to make a solution of 10 grams of salt in 100 grames of water. Include three methods to make the solution with as l

ittle time as possible.
Chemistry
2 answers:
KIM [24]2 years ago
4 0

You require distilled water, the salt and a volumetric flask

<u>Method</u>

1. Weigh accurately 10 grams of the salt using an electronic balance.

2. Add the salt to 50 ml of water in a 100ml volumetric flask and swirl until all the salt has dissolved.

3. Add distilled water up to the 100ml mark.

andreyandreev [35.5K]2 years ago
3 0

Answer:

See the explanation.

Explanation:

Hello,

This is a typical case in which an aqueous solution is prepared, therefore, we've got three ways to achieve it.

1. Use a balance to weight 10 g of the involved salt by placing it on a glass watch. Next, add the solid to an empty beaker. Then, with the beaker on it, take its reading to zero and subsequently add the water until the 100 g were measured out. Finally, shake the solution until no solid is found.

2. As mentioned before, weight the 10 g of the salt and place them on an empty beaker. Unlike the previous method, and by knowing that the density of water is 1g/mL, measure 100 mL of water using a volumetric cylinder. Finally, add the water to beaker containing the salt as shake it until no solid is found.

3. This method is quite similar to the first one, however, the beaker could be replaced by a volumetric flask.

Best regards.

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Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

Now we have to calculate the dissociated concentration of N_2O_4.

The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

Initial conc.           1.0 M          0

At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

So, the dissociate concentration of N_2O_4 = C\alpha=1.0M\times \frac{28.0}{100}=0.28M

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Now we have to calculate the concentration of N_2O_4\text{ and }NO_2 at equilibrium.

Concentration of N_2O_4 = 1.0 - x  = 1.0 - 0.28 = 0.72 M

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The expression of equilibrium constant for the reaction will be:

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Now put all the values in this expression, we get :

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8 0
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