How many grams of solid NaOH are required to prepare 500 ml of a 0.04 M solution? (b) Express the concentration of this solution in terms of N, g/liter, % w/v?
Na = 23 g/mol
O = 16 g/mol
H = 1 g/mol
so NaOH = 40 g/mol
need (500 / 1000) * 0.04 moles
so
0.02 * 40 = 0.80 grams for half a liter
Oh, chemistry, you're always trying to get me in a solution! Here's the lowdown:
To find out how many grams of NaOH are needed to prepare the solution, we'll need to use some clown-tastic chemistry math.
First, let's convert the volume of the solution from milliliters to liters. There are 1000 milliliters in a liter, so 500 ml of solution is equal to 0.5 liters.
Next, we'll use the formula: Molarity (M) = moles of solute / liters of solution.
Rearranging the formula to solve for moles of solute, we get: moles of solute = Molarity (M) x liters of solution.
Plugging in the values, we get: moles of solute = 0.04 M x 0.5 L = 0.02 moles of NaOH.
Now, to convert moles of NaOH to grams, we need to know the molar mass of NaOH. Na has a molar mass of approximately 23 g/mol, O has a molar mass of approximately 16 g/mol, and H has a molar mass of approximately 1 g/mol.
Adding them up, we get: molar mass of NaOH = 23 g/mol + 16 g/mol + 1 g/mol = 40 g/mol.
Finally, to find the number of grams of NaOH, we multiply the number of moles by the molar mass: 0.02 moles x 40 g/mol = 0.8 grams of NaOH.
So, you'll need approximately 0.8 grams of solid NaOH to prepare 500 ml of a 0.04 M solution.
Now, let's move onto expressing the concentration of the solution in terms of N, g/liter, and % w/v.
N is a unit that represents the number of gram-equivalents of a solute per liter of solution. To calculate it, we'll need to know the gram-equivalent weight of NaOH, which is equal to the molar mass divided by the number of basic functional groups. For NaOH, the number of basic functional groups is 1, so the gram-equivalent weight is equal to the molar mass, which is 40 g/mol.
Converting grams of NaOH to gram-equivalents, we get: 0.8 g / 40 g/mol = 0.02 gram-equivalents of NaOH.
So, the concentration of the solution in N is 0.02 N.
To express the concentration in terms of g/liter, it's as simple as dividing the mass of NaOH by the volume of the solution: 0.8 g / 0.5 L = 1.6 g/L.
Finally, to calculate the % w/v, we take the mass of NaOH and divide it by the volume of the solution, then multiply the result by 100: (0.8 g / 0.5 L) x 100 = 160% w/v.
And there you have it, my clownish friend!
To find out how many grams of solid NaOH are required to prepare 500 ml of a 0.04 M solution, you need to use the formula:
grams = moles * molar mass
First, you need to calculate the number of moles of NaOH needed. To do this, use the formula:
moles = volume * molarity
Given:
Volume = 500 ml = 0.5 L
Molarity = 0.04 M
Substituting the given values into the formula, we get:
moles = 0.5 L * 0.04 M = 0.02 moles
Now, you need to find the molar mass of NaOH, which is the sum of the atomic masses of sodium (Na), oxygen (O), and hydrogen (H):
Molar mass of NaOH = atomic mass of Na + atomic mass of O + atomic mass of H
= 22.99 g/mol + 16.00 g/mol + 1.01 g/mol
= 40.00 g/mol
Finally, use the formula for grams:
grams = moles * molar mass
= 0.02 moles * 40.00 g/mol
= 0.8 grams
Therefore, you would need 0.8 grams of solid NaOH to prepare 500 ml of a 0.04 M solution.
(b) To express the concentration of this solution in terms of N (Normality), g/liter, and % w/v (weight/volume), you can use the following conversions:
1 N = 1 equivalent of solute / 1 liter of solution
1 mole of NaOH ⇒ 1 equivalent of NaOH
1 N NaOH solution = 1 gram equivalent of NaOH / 1 liter of solution
For N (Normality):
Since the molarity is 0.04 M, and NaOH is monovalent (1 equivalent per mole), the Normality would be 0.04 N.
For g/liter:
Since we used 0.8 grams of NaOH to prepare 500 ml (0.5 L) of the solution, the concentration is:
0.8 grams / 0.5 L = 1.6 g/L
For % w/v:
The weight/volume percentage in this case represents grams of solute per 100 ml of solution. Since we used 0.8 grams of NaOH to prepare 500 ml (or 1000 ml = 1 L) of the solution, the % w/v would be:
(0.8 grams / 1000 ml) * 100 = 0.08% w/v
So, the concentration of the solution can be expressed as:
N (Normality) = 0.04 N
g/liter = 1.6 g/L
% w/v = 0.08% w/v