Question

Find molarity and normality and g/lit of 15v of h2o2

Answer 1

molarity = volume strength /11.2

= 15/11.2

= 1.34M

normality = volume strength /5.6

= 15/5.6

= 2.68N

♣hope it can help you ♣...........

Answer 2

Answer:

The correct answer is Normality = 2.67 N and Molarity = 1.33 M.

Let us understand about Normality and Molarity in order to solve the given question.

Explanation:

What is Normality?

The number of gram or mole equivalents of solute contained in one litre of  solution is defined as normality.

Formula and unit of Normality

• Normality = $\frac{Number of gram equivalents}{V of solution}$Number of gram equivalents / V of solution
• Number of gram equivalent =  Solute weight / Solute's equivalent weight
• N = Solute weight (g) / V x equivalent weight

Here, N = Normality and V = Volume of a solution in L.

Normality units are eqL⁻¹, N, or meqL⁻¹.

What is Molarity?

Molarity (M) is the total number of moles of a solute in 1 L of a solution. It is also known as molar concentration of a solution.

Formula and unit of Molarity

• M = Total number of moles of solute / Vol. of solution in L
• Molarity unit is mol/m³.

Relationship between Molarity and Normality

The relationship between the two can be expressed by the given formulas:

• N = M x Acidity = M x Basicity
• N = M x Molar mass / equivalent weight

(Basicity = Number of OH⁻ ions, Acidity = Number of H⁺ ions)

Now, solve the given question with the help of the above understanding.

At NTP, 1 L of H₂O₂ solution produces 15 L of O₂

                    2H₂O₂  →  2H₂O  +  O₂

                    (68 g)                     (22.4 L at NTP)

Molecular mass of H₂O₂ is 34 g. So, molecular mass of 2H₂O₂ is 32 x 2, i.e., 68 g.

H₂O₂ produces 22.4 L of O₂ at NTP = 68 g

H₂O₂ produces 15 L of O₂ at NTP = 68 / 22.4 x 15 = 45.53 g.

Strength of H₂O₂ = 45.53 g/L

Gram equivalents of H₂O₂ = $\frac{45.53}{17}$ = 2.67

(∵ gram equivalent weight of H₂O₂ = molar mass / change in oxidation number, i.e., $\frac{34}{2}$ = 17)

Now, we can easily find N and M.

Normality (N) = $\frac{2.67}{1}$ = 2.67 N

and,

Molarity (M) = $\frac{N}{2}$ = $\frac{2.67}{2}$ = 1.33 M