Question

Calculate the number of molecules of oxalic acid (H2C2O4.2H2O) in 100 mL of 0.2 N oxalic acid solution.

Answer 1

Dear Students,

● Answer -

Oxalic acid = 6.022×10^21 molecules

◆ Explanation -

# Given -

N = 0.2 N

V = 100 ml = 0.1 L

# Solution -

Normality of a solution is given by formula -

Normality = Gram equivalent of solute / volume of solution

N = E / V

E = N × V

E = 0.2 × 0.1

E = 0.02 g/mol

But gram equivalent is given by formula -

Gram equivalent = no of moles × valence

0.02 = n × 2

n = 0.02 / 2

n = 0.01 mol

No of molecules of oxalic acid present in solution can be given by -

N = n × NA

N = 0.01 × 6.022×10^23

N = 6.022×10^21 molecules

Hope I was useful..

Answer 2

1. Question⇒  Calculate the number of molecules of oxalic acid (H2C2O4.2H2O) in 100 mL of 0.2 N oxalic acid solution.

Explanation:

• # Given -
• N = 0.2 N
• V = 100 ml = 0.1 L
• # Solution -
• Normality of a solution is given by formula -
• Normality = Gram equivalent of solute / volume of solution
• N = E / V
• E = N × V
• E = 0.2 × 0.1
• E = 0.02 g/mol
• But gram equivalent is given by formula -
• Gram equivalent = no of moles × valence
• 0.02 = n × 2
• n = 0.02 / 2
• n = 0.01 mol
• No of molecules of oxalic acid present in solution can be given by -
• N = n × NA
• N = 0.01 × 6.022×10^23
• N = 6.022×10^21 molecules
• Hope I was useful..
• please mark as Brainliest