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STALIN [3.7K]
2 years ago
10

How much NaCl would you dissolve and dilute to the 100.00 mL mark in a volumetric flask with DI H2O to prepare a 0.825 M sodium

chloride solution?
Chemistry
1 answer:
adoni [48]2 years ago
4 0

Answer:

4.82 g

Explanation:

To solve this problem we'll use the <em>definition of molarity</em>:

  • Molarity = moles / liters

We are given the volume and concentration (keep in mind that 100mL=0.100L):

  • 0.825 M = moles / 0.100 L

Now we <u>calculate the number of NaCl moles required</u>:

  • moles = 0.0825 mol

Then we convert 0.0825 NaCl moles into grams, using its molar mass:

  • 0.0825 mol * 58.44 g/mol = 4.82 g
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What happens when you decrease the pressure from 150 kPa to 50 kPa on liquid water at 100 ºC?
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Suppose of ammonium nitrate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of ammonium c
Marta_Voda [28]

Answer:

Final molarity of ammonium cation in the solution = 0.16 M

Explanation:

Complete Question

Suppose 2.59 g of ammonium nitrate is dissolved in 200. mL of a 0.40M aqueous solution of sodium chromate. Calculate the final molarity of ammonium cation in the solution. You can assume the volume of the solution doesn't change when the ammonium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.

Solution

2NH₄NO₃ + Na₂CrO₄ → (NH₄)₂CrO₄ + 2NaNO₃

We first convert the given parameters to number of moles

Number of moles = (Mass/Molar mass)

Molar mass of NH₄NO₃ = 80.043 g/mol

Number of moles of NH₄NO₃ = (2.59/80.043) = 0.03224 mole

Number of moles = (Concentration in mol/L) × (Volume in L)

Number of moles of Na₂CrO₄ = 0.4 × 0.2 = 0.08 Mole

2 moles of NH₄NO₃ react with 1 mole of Na₂CrO₄

So, it it evident that NH₄NO₃ is the limiting reagent as it is in short supply in the amount needed for the reaction.

So, the number of moles of ammonium ion in the product is also 0.03224 mole.

Molarity = (Number of moles)/(Volume L)

Molarity of ammonium ion = (0.03224/0.2) = 0.1612 mol/L = 0.16 M

Hope this Helps!!!

3 0
2 years ago
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