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1 year ago

3. ml of 1.00 M LiOH to 500.00 ml of 0.075 M of LiOH (Dilution Problem)

Chemistry

1 answer:

castortr0y [4]1 year ago

6 0

The volume (in mL) of the 1.00 M LiOH needed to prepare 500.00 mL of 0.075 M of LiOH is 37.5 mL

<h3>How do I determine the volume (in mL) of LiOH needed?</h3>

Dilution formula is given as follow:

M₁V₁ = M₂V₂

Where

M₁ is the molarity of stock solution
V₁ is the volume of stock solution
M₂ is the molarity of diluted solution
V₂ is the volume of diluted solution

Applying the above formula, we obtain the volume (in mL) of LiOH needed to prepare 500.00 mL of 0.075 M of LiOH. This is illustrated below:

Molarity of stock solution of LiOH (M₁) = 1.00 M
Volume of diluted solution of LiOH (V₂) = 500.00 mL
Molarity of diluted solution of LiOH (M₂) = 0.075 M
Volume of stock solution of LiOH needed (V₁) =?

M₁V₁ = M₂V₂

1 × V₁ = 0.075 × 500

V₁ = 37.5 mL

Thus, the volume (in mL) of LiOH needed is 37.5 mL

Learn more about volume:

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4 years ago

A balloon contains 0.950 mol of nitrogen gas and has a volume of 25.5 L. How many grams of N2 should be released from the balloo

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Answer:
Mass released = 8.6 g
Explanation:
Given data:
Initial number of moles nitrogen= 0.950 mol
Initial volume = 25.5 L
Final mass of nitrogen released = ?
Final volume = 17.3 L
Solution:
Formula:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
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Initial mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Final mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - final mass
Mass released = 26.6 g - 18.0 g
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3 years ago

An unknown metal sample has a mass : of 11.29. and a volume of 2 cm3. What is the : density of the unknown metal?

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Answer:

$\boxed{\sf Density \ of \ the \ unknown \ metal = 5.645 \ g/cm^3}$

Given:

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To Find:

Density of unknown metal

Explanation:

Formula:

$\boxed{ \bold{Density = \frac{Mass}{Volume}}}$

Substituting value of mass & volume in the formula:

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3 years ago

3. ml of 1.00 M LiOH to 500.00 ml of 0.075 M of LiOH (Dilution Problem)​

3. ml of 1.00 M LiOH to 500.00 ml of 0.075 M of LiOH (Dilution Problem)