1. You have been given a sample of unknown molarity. Calculate the molarity of a solution which has been prepared by dissolving 8.75 moles of sodium chloride in enough water to produce a solution of 6.22l.
2. You have a sample which consists of 428g sodium hydroxide (NaOH) dissolved in enough water to make a final solution of 6.4l. What is the molarity?
3. You have been given a sample of 4.85 moles of glucose (C6H12O6) and you want to make a 0.75M solution with it. What will be the final volume of solution?
4. 37.0cm³ of a sodium hydroxide (NaOH) solution was pipetted into a conical flask and titrated with a standard solution of 0.45 mol dm-3 (0.3M) hydrochloric acid according to the following equation:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
A universal indicator solution was used for the titration and it was found that 22.0 cm3 of the acid (HCl) was required to neutralise the alkali (NaOH). Calculate the molarity of the sodium hydroxide and its concentration in g/dm3.
MY ANSWERS:
1. 8.75/6.22 = 875/622 = 1.406752412 = 1.41mol/L = M
2. 428/6.41 = 42800/641 = 66.770670826833... = 66.77mol/L = M
3. 4.85/0.75 = 97/15 = 6.46 = 6.46mol/L = M
QUESTION 4 I HAVE NO CLUE PLEASE PLEASE SOMEBODY HELP ME
Your answers for 1 and 3 are correct. The answer for question 2 contains an error in the calculation. Here are the revised solutions:
1. This calculation is correct: 8.75 moles / 6.22L = 1.41 mol/L.
2. First, calculate the moles of NaOH: 428g / (40 g/mol) = 10.7 moles. Then, divide by the volume of the solution: 10.7 moles / 6.4 L = 1.67 mol/L.
3. Your calculation is right: 4.85 moles / 0.75 M = 6.46 L.
4. To find the molarity of sodium hydroxide, first, we need to find the moles of HCl used in the titration.
Moles of HCl = (0.3 mol/dm³) * (22.0 cm³) / (1000 cm³/dm³) = 0.0066 moles
Since the balanced equation indicates one mole of NaOH reacts with one mole of HCl, there must have been 0.0066 moles of NaOH in the 37 cm³ of NaOH solution.
Moles of NaOH / Volume of NaOH solution = Molarity of NaOH
0.0066 moles / (37 cm³ / 1000 cm³/dm³) = 0.1784 mol/dm³ = 0.178 mol/L
To find the concentration of NaOH in g/dm3, multiply the molarity by the molar mass of NaOH:
0.178 mol/L * 40 g/mol = 7.12 g/dm³
To find the molarity of the sodium hydroxide (NaOH) solution in question 4, you can use the equation M1V1 = M2V2, where M1 is the molarity of the hydrochloric acid (HCl) solution, V1 is the volume of HCl used in the titration, M2 is the molarity of NaOH solution, and V2 is the volume of NaOH used in the titration.
Given:
M1 = 0.45 mol/dm³ (0.3M)
V1 = 22.0 cm³
V2 = 37.0 cm³
First, convert the volumes to dm³:
V1 = 22.0 cm³ = 22.0/1000 dm³ = 0.022 dm³
V2 = 37.0 cm³ = 37.0/1000 dm³ = 0.037 dm³
Now, substitute the values into the equation:
M1V1 = M2V2
(0.45 mol/dm³)(0.022 dm³) = M2(0.037 dm³)
Simplify and solve for M2:
0.0099 mol = M2(0.037 dm³)
M2 = 0.0099 mol / 0.037 dm³
M2 = 0.267 mol/dm³ = 0.267M
Therefore, the molarity of the sodium hydroxide solution is 0.267 mol/dm³ (0.267M).
To calculate the concentration in g/dm³, you can use the molar mass of NaOH (40.00 g/mol) and the molarity (0.267 mol/dm³):
Concentration (g/dm³) = Molarity (mol/dm³) x Molar mass (g/mol)
Concentration = 0.267 mol/dm³ x 40.00 g/mol
Concentration = 10.68 g/dm³
Therefore, the concentration of the sodium hydroxide solution is 10.68 g/dm³.
To find the molarity of the sodium hydroxide (NaOH) solution in question 4, you can use the equation and the volume of the acid (HCl) required for neutralization.
Step 1: Calculate the number of moles of HCl used:
Moles of HCl = molarity × volume in liters
Moles of HCl = 0.45 mol/dm³ × 0.022 dm³ (converted cm³ to dm³)
Moles of HCl = 0.0099 mol
Step 2: Since the balanced equation shows a 1:1 ratio between NaOH and HCl, the number of moles of NaOH is also 0.0099 mol.
Step 3: Calculate the molarity of NaOH:
Molarity of NaOH = moles of NaOH / volume in liters
Molarity of NaOH = 0.0099 mol / 0.037 dm³ (converted cm³ to dm³)
Molarity of NaOH = 0.267 mol/dm³ = 0.267 M
Step 4: Calculate the concentration in grams/dm³:
Concentration in g/dm³ = molarity × molar mass of NaOH
Concentration in g/dm³ = 0.267 mol/dm³ × 40 g/mol
Concentration in g/dm³ = 10.68 g/dm³ = 10.68 g/L
Therefore, the molarity of the sodium hydroxide solution is 0.267 M, and its concentration is 10.68 g/dm³.