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butalik [34]
2 years ago
12

You need to prepare 500 mL of a 0.250 M NaOH solution for a titration. What mass of NaOH is needed to

Chemistry
1 answer:
VikaD [51]2 years ago
5 0

The mass of NaOH needed to prepare the solution is 5 g

<h3>What is molarity? </h3>

This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

<h3>How to determine the mole of sodium hydroxide, NaOH</h3>

We'll begin by calculating the moles of NaOH. This can be obtained as follow:

  • Volume = 500 mL = 500 / 1000 = 0.5 L
  • Molarity = 0.25 M
  • Mole of NaOH =?

Mole = Molarity x Volume

Mole of NaOH = 0.25 × 0.5

Mole of NaOH = 0.125 mole

<h3>How to determine the mass of sodium hydroxide, NaOH</h3>

The mass of NaOH needed can be obtained as follow:

  • Molar mass of NaOH = 40 g/mol
  • Mole of NaOH = 0.125 mole
  • Mass of NaOH =?

Mass = mole × molar mass

Mass of NaOH = 0.125 × 40

Mass of NaOH = 5 g

Thus, to prepare the solution, measure 5 g of NaOH and place it in a 500 mL volumetric flask and make it up to the mark with water.

Learn more about molarity:

brainly.com/question/15370276

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3 years ago
Balance this equation Hgo —&gt; Hg + O2
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6 0
3 years ago
Oftentimes solubility of a compound limits the concentration of the solution that can be prepared. Use the solubility data given
bulgar [2K]

Answer:

NaNO3 (solubility = 89.0 g/100 g H2O)

Explanation:

The solubility of a specie is the amount of solute that will dissolve in one litre of the solvent. Solubility is usually expressed in units of molarity.

Now let us calculate the molarity of the NaNO3 (solubility = 89.0 g/100 g H2O)

Molar mass of NaNO3= 23+14+3(16)= 85gmol-1

Mass of solute=89.0g

Amount of solute= mass of NaNO3/molar mass of NaNO3

Amount of solute= 89.0g/85.0 gmol-1

= 1.0moles of NaNO3

Note that 100g of water=100cm^3 of water.

If 1.0 moles of NaNO3 dissolve in 100cm^3 or water therefore,

x moles of NaNO3 will dissolve in 1000cm^3 of water

x= 1.0 × 1000/ 100

x= 10.0 moles of NaNO3

3 0
4 years ago
I really need help on this question.
natita [175]
I need that answer too
7 0
3 years ago
How many moles of hypomanganous acid. H3 MnO4, are contained in 22.912 g?
gregori [183]

Answer:

0.188mol

Explanation:

Using the formula;

mole = mass/molar mass

Molar mass of hypomanganous acid. (H3MnO4) = 1(3) + 55 + 16(4)

= 3 + 55 + 64

= 122g/mol

According to this question, there are 22.912g of H3MnO4

mole = 22.912g ÷ 122g/mol

mole = 0.188mol

6 0
3 years ago
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