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3 years ago

To what volume would you need to dilute 200 mL of a 5.85M solution of Ca(OH)2 to make it a 1.95M solution?

Chemistry

1 answer:

Blizzard [7]3 years ago

8 0

Answer: 600 mL

Explanation:

Given that;

M₁ = 5.85 m

M₂ = 1.95 m

V₁ = 200 mL

V₂ = ?

Now from the dilution law;

M₁V₁ = M₂V₂

so we substitute

5.85 × 200 = 1.95 × V₂

1170 = 1.95V₂

V₂ = 1170 / 1.95

V₂ = 600 mL

Therefore final volume is 600 mL

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Which solution has the higher boiling point?

dedylja [7]

NaCl has higher boiling point than C₂H₆O₂ in H₂O.

<h3>What is molality?</h3>

The number of moles of solute in a solution equal to 1 kg or 1000 g of solvent is referred to as its molality.
The definition of molarity, on the other hand, is based on a certain volume of solution.
Mol/kg is a typical molality measurement unit in chemistry.

<h3>Calculation of boiling point:</h3>

When the non-volatile solute is dissolved in a solvent, the boiling point rises along with the molality (concentration) of the solute.

Given,

Mass of C₂H₆O₂ (solute) = 15.0 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of C₂H₆O₂ = 62 g/mol

Thus, the moles of solute = 15.0 g x 1.0 mol solute/62 g/mol

= 0.2419 mol

Therefore, molality(m) of C₂H₆O₂ (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2419/0.50

= 0.4838 m

Similarly,

Given ,

Mass of NaCl (solute) = 15 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of NaCl (solute) = 58.44 g/mol

Thus, the moles of NaCl (solute) = 15.0 g x 1.0 mol solute/58.44

= 0.2566 mol

Therefore, molality(m) of NaCl (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2566 mol/0.50 kg

= 0.51 m

Hence, the molality of NaCl (solute) is more than the molality of C₂H₆O₂(solute).

So, with an increase in the solute's concentration (molality), the boiling point rises.

Therefore, NaCl has higher boiling point than C₂H₆O₂ in H₂O.

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What kind of molecule is glucose

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Organic molecule. In particular, a carbohydrate.

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3 years ago

Calculate the number of atoms in 0.25 moles of manganese?

White raven [17]

<h3>Answer:</h3>

1.5 × 10²³ atoms Mn

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.25 mol Mn

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 0.25 \ mol \ Mn(\frac{6.022 \cdot 10^{23} \ atoms \ Mn}{1 \ mol \ Mn})$
Multiply: $\displaystyle 1.5055 \cdot 10^{23} \ atoms \ Mn$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.5055 × 10²³ atoms Mn ≈ 1.5 × 10²³ atoms Mn

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