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antiseptic1488 [7]
3 years ago
11

Calculate the weight in grams of the primary standard solid oxalic acid needed to prepare 250 ml of a 0.200 n solution to be use

d in this laboratory exercise.
Chemistry
1 answer:
slamgirl [31]3 years ago
3 0

Answer:

2.25 g.

Explanation:

  • Firstly, we can define the noramlity as <em>the number of gram equivalents of the solute in 1.0 L of the solution.</em>

<em>N = (mass / equivalent mass) of the solute x (1000 / V of the solution)</em>

The solute is oxalic acid.

N is the normality of the solution <em>(N = 0.2 N)</em>,

mass of the solute (oxalic acid) (??? needed to be calculated),

equivalent mass of the oxalic acid = molar mass / n (the no. of reproducible H) = (90.03 g/mol) / (2) = 45.015.

V is the volume of the solution (V = 250 ml).

∴ The mass of the oxalic acid in g = (N x equivalent mass x V) / 1000 = (0.2 N) (45.015) (250.0 ml) / 1000 = 2.25 g.

<u><em>Very important note:</em></u>

If the solute source is oxalic acid dihydrate:

The molar mass = 126.07 g/mol.

The equivalent mass = 63.035 g.equ./mol.

∴ The mass of the oxalic acid in g = (N x equivalent mass x V) / 1000 = (0.2 N) (63.035) (250.0 ml) / 1000 = 3.15 g.

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7. If the pOH of an RbOH solution is 6.32, what is the concentration (molarity) of the base?
liubo4ka [24]

4.79 x 10⁻⁷moldm⁻³

Explanation:

Given parameters:

    pOH of RbOH = 6.32

Unknown:

Molarity of the base = ?

Solution:        

    The pH or pOH scale is used for expressing the level of acidity alkalinity of aqueous solutions.

                      pOH = -log[OH⁻]

  we know the pOH to be 6.32

               6.32 = -log[OH⁻]

               [OH⁻]  = inverse log₁₀(6.32)

                [OH⁻] =  4.79 x 10⁻⁷moldm⁻³

Learn more:

pH brainly.com/question/12985875

#learnwithBrainly

6 0
3 years ago
As the tempreture of a liquid increases the solubility of a liquid of that liquid
anzhelika [568]
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I hope this helps and also I assumed that your question involved the solubility of an ionic substance in a solvent like water.  If that was not your question feel free to say so in the comments so that I can answer your actually question.
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Explanation:

7 0
2 years ago
The osmotic pressure of a solution formed by dissolving 35.0 mg of aspirin (c9h8o4) in 0.250 l of water at 25°c is __________ at
Lunna [17]
Mass of aspirin = 0.025 g
Molar mass of C9H8O4 is 180.1583 g/mol
moles of aspirin = .025g / 180.1583 g/mol = 0.000138767 moles
volume solution = .250 L
molarity of the solution = 0.000138767 moles / .250L =5.551 x 10 ^-04 Moles / liter
for aspirin i = Vant'Hoff factor = 1 particle in solution
T = 25 + 273 =298 K
osmotic pressure = M x R x T x i =
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6 0
3 years ago
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