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.
1. To calculate the molarity of the sodium chloride solution, you divide the number of moles of sodium chloride (8.75 moles) by the volume of the solution (6.22 L). That gives us a molarity of approximately 1.41 M.
2. For the sodium hydroxide solution, we divide the number of grams of sodium hydroxide (428 g) by the molar mass of sodium hydroxide (22.99 g/mol + 16.00 g/mol + 1.01 g/mol). Then, divide that result by the volume of the solution in liters (6.4 L). This gives us a molarity of approximately 10.04 M.
3. To determine the final volume of the 0.75 M glucose solution, we divide the number of moles of glucose (4.85 moles) by the desired molarity (0.75 M). This gives us a volume of approximately 6.47 L.
4. In the titration between sodium hydroxide and hydrochloric acid, we can use the following equation to find the molarity of sodium hydroxide:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Since 22.0 mL of 0.45 M hydrochloric acid is required to neutralize the sodium hydroxide, we can set up a proportion:
[0.45 M HCl] / (22.0 mL) = [x M NaOH] / (37.0 mL)
Solving for x, we find that the molarity of the sodium hydroxide is approximately 0.78 M.
To calculate its concentration in g/dm3, we need to convert the molarity to moles per liter. The molar mass of sodium hydroxide (NaOH) is 22.99 g/mol + 16.00 g/mol + 1.01 g/mol, which equals 40.00 g/mol. Therefore, the concentration of sodium hydroxide in g/dm3 is approximately 31.2 g/dm3.
1. To calculate the molarity of the solution, we need to use the formula:
Molarity (M) = moles of solute / volume of solution in liters
Given:
moles of sodium chloride (NaCl) = 8.75 moles
volume of solution = 6.22 L
Molarity = 8.75 moles / 6.22 L ≈ 1.409 M
The molarity of the solution is approximately 1.409 M.
2. To calculate the molarity, we need to use the formula:
Molarity (M) = moles of solute / volume of solution in liters
Given:
mass of sodium hydroxide (NaOH) = 428 g
volume of solution = 6.4 L
First, we need to convert the mass of NaOH to moles:
moles of NaOH = mass (g) / molar mass (g/mol)
Molar mass of NaOH = 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 39.99 g/mol
moles of NaOH = 428 g / 39.99 g/mol ≈ 10.70 moles
Molarity = 10.70 moles / 6.4 L ≈ 1.672 M
The molarity of the solution is approximately 1.672 M.
3. To calculate the final volume of the solution, we can use the formula:
Molarity (M) = moles of solute / volume of solution in liters
Given:
moles of glucose (C6H12O6) = 4.85 moles
desired molarity = 0.75 M
Rearranging the formula, we get:
Volume of solution (L) = moles of solute / Molarity
Volume of solution = 4.85 moles / 0.75 M ≈ 6.47 L
The final volume of the solution is approximately 6.47 L.
4. To calculate the molarity of the sodium hydroxide (NaOH) solution, we can use the equation for the reaction:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Given:
volume of NaOH solution = 37.0 cm³
volume of HCl solution required for neutralization = 22.0 cm³
molarity of HCl = 0.45 mol/dm³ (or 0.3 M)
We can use the balanced equation to determine the moles of NaOH:
1 mole of NaOH reacts with 1 mole of HCl
moles of NaOH = moles of HCl = molarity of HCl x volume of HCl solution (in liters)
= 0.45 mol/dm³ x (22.0 cm³ / 1000 cm³/dm³)
= 0.0099 moles
Now, we can calculate the molarity of the NaOH solution:
Molarity = moles of NaOH / volume of NaOH solution (in liters)
= 0.0099 moles / (37.0 cm³ / 1000 cm³/L)
≈ 0.267 M
The molarity of the sodium hydroxide solution is approximately 0.267 M.
To calculate the concentration of the sodium hydroxide solution in g/dm³, we need to convert moles to grams:
mass of NaOH = moles of NaOH x molar mass of NaOH
Molar mass of NaOH = 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 39.99 g/mol
mass of NaOH = 0.0099 moles x 39.99 g/mol ≈ 0.40 g
Concentration = mass of NaOH / volume of NaOH solution (in liters)
= 0.40 g / (37.0 cm³ / 1000 cm³/L)
≈ 10.81 g/dm³
The concentration of the sodium hydroxide solution is approximately 10.81 g/dm³.
1. To calculate the molarity of a solution, you need to divide the number of moles of solute by the volume of the solution in liters.
Given:
- Moles of sodium chloride (NaCl): 8.75 moles
- Volume of solution: 6.22 L
Molarity (M) = Moles of solute / Volume of solution
Molarity (NaCl) = 8.75 moles / 6.22 L
Molarity (NaCl) ≈ 1.41 M
Therefore, the molarity of the solution prepared by dissolving 8.75 moles of sodium chloride in 6.22 liters of water is approximately 1.41 M.
2. Similarly, to calculate the molarity of a solution, you need to divide the number of moles of solute by the volume of the solution in liters.
Given:
- Mass of sodium hydroxide (NaOH): 428 g
- Volume of solution: 6.4 L
First, convert the mass of NaOH to moles using its molar mass:
Molar mass of NaOH = 22.99 g/mol (from Na) + 16.00 g/mol (from O) + 1.01 g/mol (from H) = 39.00 g/mol
Moles of NaOH = Mass of NaOH / Molar mass of NaOH
Moles of NaOH = 428 g / 39.00 g/mol ≈ 10.97 moles
Molarity (NaOH) = Moles of solute / Volume of solution
Molarity (NaOH) = 10.97 moles / 6.4 L
Molarity (NaOH) ≈ 1.72 M
Therefore, the molarity of the solution prepared by dissolving 428 g of sodium hydroxide in 6.4 liters of water is approximately 1.72 M.
3. To calculate the final volume of a solution, you can use the formula:
Molarity (M) = Moles of solute / Volume of solution
Given:
- Moles of glucose (C6H12O6): 4.85 moles
- Molarity of solution: 0.75 M
Rearrange the formula to solve for the volume of solution:
Volume of solution = Moles of solute / Molarity
Volume of solution = 4.85 moles / 0.75 M
Volume of solution ≈ 6.47 L
Therefore, to make a 0.75 M solution using 4.85 moles of glucose, the final volume of the solution should be approximately 6.47 liters.
4. To calculate the molarity of the sodium hydroxide (NaOH) solution, you can use the equation:
Molarity (NaOH) x Volume (NaOH) = Molarity (HCl) x Volume (HCl)
Given:
- Volume of NaOH solution: 37.0 cm³
- Volume of HCl solution: 22.0 cm³
- Molarity of HCl solution: 0.45 mol/dm³ (or 0.3 M, since they are equivalent)
Rearrange the equation to solve for the molarity of NaOH:
Molarity (NaOH) = (Molarity (HCl) x Volume (HCl)) / Volume (NaOH)
Molarity (NaOH) = (0.45 mol/dm³ x 22.0 cm³) / 37.0 cm³
Molarity (NaOH) ≈ 0.27 mol/dm³ (or 0.18 M)
To calculate the concentration of NaOH in g/dm³, multiply the molarity by the molar mass of NaOH:
Concentration (NaOH) = Molarity (NaOH) x Molar mass (NaOH)
Molar mass of NaOH = 22.99 g/mol (from Na) + 16.00 g/mol (from O) + 1.01 g/mol (from H) = 39.00 g/mol
Concentration (NaOH) = 0.27 mol/dm³ x 39.00 g/mol
Concentration (NaOH) ≈ 10.53 g/dm³