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1 year ago

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]1 year 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:

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What are the formulas for the ionic compounds named sodium iodide and magnesium iodide

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Sodium Iodide: NaI

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3 years ago

15.1 L N2 at 25 °C and 125 kPa and 44.3 L O2 at 25 °C and 125 kPa were transferred to a tank with a volume of 6.25 L. What is th

Y_Kistochka [10]

Answer:

The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

Explanation:

For N2,

Pressure(P₁)=125 kPa

Volume(V₁)=15·1 L

Temperature (T₁)=25°C=25+273 K=298 K

Similarly, for Oxygen,

Pressure(P₂)= 125 kPa

Volume(V₂)= 44.3 L

Temperature(T₂)=25°C= 298 K

Then, for the mixture,

Volumeof the mixture( V)= 6.25 L

Pressure(P)=?

Temperature (T)= 51°C = 51+273 K=324 K

Then, By Combined gas laws,

$\frac{P_{1} V_{1} }{T_{1} } +\frac{P_{2} V_{2} }{T_{2} } =\frac{PV}{T}$

or, $\frac{15.1*125}{298} +\frac{44.3*125}{298} =\frac{P*6.25}{324}$

or, $6.34+18.58=\frac{P*6.25}{324}$

or, $P=\frac{24.92*324}{6.25}$

∴P=1291.85 kPa

So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

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3 years ago

How many sodium ions are in 3.0 moles of nacl

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1 molecule of NaCl contains 1 sodium ion (Na+), that's why if we have 3.0 moles of.
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3 years ago

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Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

= 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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3 years ago

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Answer:

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3 years ago

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