calculate the normality of 4g of sodium hydroxide in 100ml of solution

To calculate the normality of a solution, you need to know the molarity and the number of equivalents of the solute.

1. Start by calculating the molarity of sodium hydroxide (NaOH) in the solution:
Molarity (M) = moles of solute / volume of solution (in liters)

Given that you have 4 grams of NaOH in 100 mL of solution, you first need to convert the volume to liters:
Volume = 100 mL = 100/1000 L = 0.1 L

The molar mass of NaOH is:
Na = 22.99 g/mol
O = 16 g/mol
H = 1.01 g/mol

Molar mass of NaOH = 22.99 + 16 + 1.01 = 39.01 g/mol

Now, convert the mass of NaOH to moles:
Moles = mass / molar mass = 4 g / 39.01 g/mol

Next, divide the moles by the volume to find the molarity:
Molarity (M) = moles / volume = (4 g / 39.01 g/mol) / 0.1 L

2. Calculate the number of equivalents of NaOH:
Since NaOH is a base, it dissociates into one hydroxide ion (OH-) and one sodium ion (Na+). Both ions have a charge of 1, which means the equivalent weight is the same as the molar weight.

Therefore, the number of equivalents = molar mass = 39.01 g/mol

3. Calculate the normality by multiplying molarity by the number of equivalents:
Normality = Molarity * number of equivalents

In this case, the normality would be:
Normality = (4 g / 39.01 g/mol) / 0.1 L * 39.01 g/mol

Simplifying further:
Normality = 4 / 0.1
Normality = 40 N

So, the normality of the solution containing 4g of sodium hydroxide in 100ml of solution is 40N.

equivalents = grams/eq weight = 4/40 = ?

N = eq/L = ?

Equivalent weight=Mw÷valence

=40÷2
20g
1N:20g:1000ml
?:4g:100ml
So 1N:20g
?:4g
?=0.2N
Therefore 0,2N =2N