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?
ALLAH HU AKBAR
0.02 moles is required=0.5L*0.04 M
500mL=0.5L
0.02 moles*(39.997g/mole)=???
(???/500)*100= w/v %
Only 1 equivalent. So, M is equal to N
g/L= ???/0.5L
Answer should have only one significant figure, but you can include 3.
To find the number of grams of solid NaOH required to prepare a 0.04 M solution, you need to know the molar mass of NaOH and the volume of the solution.
The molar mass of NaOH is calculated by adding the atomic masses of its constituent elements: sodium (Na), oxygen (O), and hydrogen (H). The atomic masses are: Na = 22.99 g/mol, O = 16.00 g/mol, and H = 1.01 g/mol.
So the molar mass of NaOH is calculated as follows:
NaOH = (Na: 1 * 22.99) + (O: 1 * 16.00) + (H: 1 * 1.01) = 39.99 g/mol
Now, let's calculate the number of moles of NaOH required:
moles = Molarity * Volume (in liters)
Given that the volume is 500 ml (which is 0.5 liters) and the molarity is 0.04 M, the number of moles is:
moles = 0.04 * 0.5 = 0.02 moles
To calculate the number of grams, multiply the number of moles by the molar mass:
grams = moles * molar mass
grams = 0.02 * 39.99 = 0.7998 grams
Therefore, approximately 0.8 grams of solid NaOH is required to prepare 500 ml of a 0.04 M solution.
b) Expressing the concentration of the solution in terms of N, g/liter, and % w/v:
N is a unit of concentration representing the number of equivalents of solute per liter of solution. In the case of NaOH, as it is a monoprotic base, 1 mole of NaOH reacts with 1 mole of H+ ions, so the equivalent weight is equal to the molar mass of NaOH (39.99 g/mol).
To express the concentration in terms of N, we can use the formula:
N = Molarity * Equivalent Weight
N = 0.04 * 39.99 = 1.5996 N
Therefore, the concentration of the solution in terms of N is approximately 1.6 N.
To express the concentration in g/liter, you can convert the molarity using the molar mass:
Concentration in g/liter = Molarity * molar mass = 0.04 * 39.99 = 1.5996 g/liter
Therefore, the concentration of the solution in terms of grams per liter is approximately 1.6 g/liter.
To express the concentration as a percentage (w/v), you need to know the weight of NaOH in grams and the volume of the solution in milliliters:
% w/v = (Weight of NaOH in grams / Volume of solution in milliliters) * 100
Given that we already calculated the weight of NaOH as 0.8 grams and the volume of the solution is 500 ml, the percentage w/v is:
%w/v = (0.8 / 500) * 100 = 0.16%
Therefore, the concentration of the solution in terms of % w/v is approximately 0.16% w/v.