What is the molarity of a 2.0 L sodium hydroxide solution containing 10.0 grams of solute?

1 answer:

8 0

Molar mass NaOH ( sodium hydroxide) = 40.0 g/mol

Number of moles:

n = mass / molar mass

n = 10.0 / 40.0

n = 0.25 moles

Volume = 2.0 L

Molarity = number of moles / volume in liters

M = 0.25 / 2.0

M = 0.125 M

answer D

hope this helps!

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<h3>Answer:</h3>

23.459 g NaNO₂

<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>Stoichiometry</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] H₂SO₄ + 2NaNO₂ → 2HNO₂ + Na₂SO₄

[Given] 24.14714 g Na₂SO₄

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

[RxN] 1 mol Na₂SO₄ = 2 mol NaNO₂

Molar Mass of Na - 22.99 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of Na₂SO₄ - 2(22.99) + 32.07 + 4(16.00) = 142.05 g/mol

Molar Mass of NaNO₂ - 22.99 + 14.01 + 2(16.00) = 69.00 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 24.14714 \ g \ Na_2SO_4(\frac{1 \ mol \ Na_2SO_4}{142.05 \ g \ Na_2SO_4})(\frac{2 \ mol \ NaNO_2}{1 \ mol \ Na_2SO_4})(\frac{69.00 \ g \ NaNO_2}{1 \ mol \ NaNO_2})$
Multiply/Divide: $\displaystyle 23.4587 \ g \ NaNO_2$