To determine the normality of a 500 ml solution of sodium hydroxide (NaOH) containing 30 grams of NaOH, follow the steps below:
1. Convert the volume of the solution from milliliters to liters:
The given volume is 500 ml. We know that 1000 ml equals 1 liter, so:
[tex]\[
\text{Volume in liters} = \frac{500 \text{ ml}}{1000 \text{ ml/L}} = 0.5 \text{ L}
\][/tex]
2. Calculate the number of moles of NaOH:
The molecular weight of NaOH is given as 40 g/mol (Na: 23 g/mol, O: 16 g/mol, H: 1 g/mol).
Given mass is 30 grams. Using the molecular weight, we can determine the number of moles:
[tex]\[
\text{Moles of NaOH} = \frac{\text{mass of NaOH}}{\text{molecular weight of NaOH}} = \frac{30 \text{ g}}{40 \text{ g/mol}} = 0.75 \text{ mol}
\][/tex]
3. Calculate the normality of the solution:
Normality (N) measures the concentration of equivalents per liter of solution. Since NaOH has one replaceable OH⁻ ion, its normality is the same as its molarity in this case.
To find the normality, use the formula:
[tex]\[
\text{Normality (N)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.75 \text{ moles}}{0.5 \text{ liters}} = 1.5 \text{ N}
\][/tex]
Therefore, the normality of the 500 ml solution of sodium hydroxide containing 30 grams of NaOH is [tex]\(1.5 \, \text{N} \)[/tex].
