(1)Make 500 ML of 4N H3PO4. How many grams will be needed?
(2) In a titration procedure, 20mL of 2N solution were required to titrate 5mL of an unknown solution. What is the normality of the solution?
(3) How much of a 0.01M can be made from 20mL of 6M HCl?
(4)How many grams of NaOH are present in 300mL of a 6M solution of sodium hydroxide?
you want 2 mols (half a liter of 4N)
H3PO4 = 3+31+4(16) grams/mol
molarity=gramsSolute/(MolMass*Liters)
a) gramsSolute=Molarity*Molmass*Liters. I have no idea what ML means, do you mean millilters, or 1/2 Liter?
b) Normality=Heq*Molarity
VaNa=Vb*Nb
Nb=VaNa/Vb=.020*2/.005=8N
c. you are diluting it 6/.01 or 600 times, or 60O*20ml=18Liters
d. massSolute=Molarity/(volumeL*molmass)
= 6/(.3*40)
I think this question should be reviewed and/or rethought.
a. mL x N x milliequivalent weight = grams.
500 x 4 x (98/3000) = about 65 g.
b. 8N is right
c. There is a typo here. 600*20 = 12,000 mL or 12 L.
d. mols = M x L = 6 x 0.3 = 1.8 mols. Then grams = mols x molar mass = 1.8 x 40 = 72 g NaOH OR
g = M x L x molar mass = 6 x 0.3 x 40 = about 72 grams.
(1) To find the number of grams needed to make 500 mL of a 4N H3PO4 solution, we need to use the formula:
N = (grams / molar mass) / volume (in liters)
In this case, N is given as 4, volume is 500 mL (which is equal to 0.5 L), and molar mass of H3PO4 is 97.994 g/mol.
We can rearrange the formula to solve for grams:
grams = N * molar mass * volume
Substituting the given values, we get:
grams = 4 * 97.994 g/mol * 0.5 L = 195.988 g
Therefore, you will need 195.988 grams of H3PO4 to make a 500 mL 4N solution.
(2) To find the normality of the solution, we can use the formula:
Normality (N1) * volume (V1) = Normality (N2) * volume (V2)
In this case, we have:
N1 = 2N (given normality)
V1 = 20 mL (volume of the 2N solution)
N2 = unknown normality
V2 = 5 mL (volume of the unknown solution)
Substituting the values into the formula:
2N * 20 mL = N2 * 5 mL
Rearranging the formula, we can solve for N2:
N2 = (2N * 20 mL) / 5 mL = 8N
Therefore, the normality of the unknown solution is 8N.
(3) To find the amount of 0.01M solution that can be made from 20 mL of 6M HCl, we need to use the formula:
M1 * V1 = M2 * V2.
In this case, we have:
M1 = 6M (molarity of the 6M HCl solution)
V1 = 20 mL (volume of the 6M HCl solution)
M2 = 0.01M (desired molarity)
V2 = unknown (volume of the 0.01M solution to be made)
Substituting the values into the formula:
6M * 20 mL = 0.01M * V2
Solving for V2:
V2 = (6M * 20 mL) / 0.01M = 12000 mL
Therefore, you can make 12000 mL (or 12 L) of a 0.01M solution from 20 mL of 6M HCl.
(4) To find the number of grams of NaOH present in 300 mL of a 6M solution of sodium hydroxide, we can use the formula:
moles = molarity * volume (in liters)
In this case, molarity is given as 6M and the volume is 300 mL (which is equal to 0.3 L).
Plugging in the values into the formula, we get:
moles = 6M * 0.3 L = 1.8 moles
To find the grams, we need to multiply the moles by the molar mass of NaOH, which is 39.997 g/mol. Therefore:
grams = 1.8 moles * 39.997 g/mol = 71.9946 grams
Therefore, there are 71.9946 grams of NaOH present in 300 mL of a 6M solution of sodium hydroxide.