2. How many grams of 96.5% pure NaOH pellets do you need to make 1.000 liter of 1.50 N NaOH?
To determine the number of grams of 96.5% pure NaOH pellets needed to make 1.000 liter of 1.50 N NaOH, you need to know the molecular weight of NaOH and perform some calculations.
First, you need to find the molecular weight of NaOH, which is comprised of one sodium atom (Na), one oxygen atom (O), and one hydrogen atom (H). The atomic weights of sodium, oxygen, and hydrogen are approximately 22.99 g/mol, 16 g/mol, and 1 g/mol, respectively.
The total molecular weight of NaOH can be calculated by adding the atomic weights of its constituent atoms:
Na = 22.99 g/mol
O = 16 g/mol
H = 1 g/mol
Total molecular weight of NaOH = 22.99 + 16 + 1 = 39.99 g/mol
Now, you can calculate the number of moles of NaOH needed to prepare 1.000 liter of 1.50 N NaOH. To do this, you need to use the formula:
Moles = Normality × Volume (in liters)
In this case, the given normality of NaOH is 1.50 N, and the volume is 1.000 L. Plugging in these values, we get:
Moles = 1.50 N × 1.000 L = 1.50 moles
Since you now have the number of moles of NaOH needed, you can calculate the mass of NaOH required. Simply multiply the moles of NaOH by the molecular weight of NaOH:
Mass (in grams) = Moles × Molecular weight
Mass = 1.50 moles × 39.99 g/mol ≈ 59.99 grams
Therefore, you would need approximately 59.99 grams of 96.5% pure NaOH pellets to make 1.000 liter of 1.50 N NaOH.