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lord [1]
3 years ago
7

56.5 g of sodium nitrate is dissolved in water to make 627 g of solution . what is the percent sodium nitrate in the solution

Chemistry
2 answers:
Tamiku [17]3 years ago
8 0

Answer:

9.01%

Explanation:

The following data were obtained from the question:

Mass of sodium nitrate, NaNO3 = 56.5g

Mass of solution = 627g

The percentage composition of sodium nitrate, NaNO3 in the solution can be obtained as follow:

Percentage composition of NaNO3 = Mass of NaNO3/mass of solution x 100

Percentage composition of NaNO3 = 56.5/627 x 100 = 9.01%

Therefore, the percentage composition of sodium nitrate, NaNO3 in the solution is 9.01%

Brrunno [24]3 years ago
8 0

Answer:

The percentage of sodium nitrate in the solution is 9.01%

Explanation:

Here we have;

56.5 g of sodium nitrate dissolved in water to make a solution of mass = 627 kg

Mass of sodium nitrate, NaNO₃, in the solution = 56.5 g

Mass of resultant solution = 627 g

The relation for the percentage of a solute by mass dissolved in a solvent is presented as follows;

Percentage \, by  \, mass = \frac{Mass \, of \, solute}{Mass \, of \, solution} \times 100

Therefor, for the 56.5 g of sodium nitrate solution, we have;

Percentage \, by  \, mass \, of \, NaNO_3 = \frac{Mass \, of \, NaNO_3}{Mass \, of \, solution} \times 100 = \frac{56.5 \, g}{627 \, g} \times 100 = 9.01 \%

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Answer:

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Explanation:

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∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

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∆U = Cv((45°C) - (25°C))

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The Initial mole is calculated as

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So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

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R = 8.314J/molK

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n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

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