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.
The first three questions may be solved using the relation:
molarity (mols)
= number of moles / volume (L)
where
number of moles
= mass (g) / molar mass (g).
Example:
10.0 g of NaCl is dissolved to make 0.523 L of solution. Find molarity of the NaCl solution.
number of moles
= 10/(23+35.5)=0.171
molarity
=0.171/0.523
=0.327
For #4,
From the equation, it can be deduced that neutralization requires equal moles of acid and base.
=>
#of moles used for acid and base are equal.
Use the above relation to solve for # of moles.
Note: the molarity of HCl is not clear. 0.45 mol-dm-3≡0.45 mol/L but does NOT equal 0.3M.
1. Well, if we dissolve 8.75 moles of sodium chloride in enough water to produce a solution of 6.22 liters, we can call it a "salty surprise!" But to calculate the molarity, we can use the formula Molarity (M) = moles of solute / liters of solution. So in this case, Molarity = 8.75 moles / 6.22 liters.
2. Ah, sodium hydroxide, the superhero of cleaning products! To find the molarity, we can use the same formula: Molarity (M) = moles of solute / liters of solution. So, for 428 grams of NaOH dissolved in 6.4 liters, we'll need to convert grams to moles first and then divide by the volume in liters.
3. So, you have 4.85 moles of glucose and you want to make a 0.75M solution with it. It's like playing mix-and-match but with chemistry! To find the final volume of the solution, we can use the formula Molarity (M) = moles of solute / liters of solution. Rearranging the equation, we have liters of solution = moles of solute / Molarity.
4. Let's get our titration on! You've pipetted 37.0 cm³ of NaOH solution and titrated it with 0.45 mol dm-³ HCl. According to the equation NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l), we know that the ratio is 1:1. So, if 22.0 cm³ of HCl is needed for neutralization, we can say that 22.0 cm³ of HCl is equal to 0.45 mol dm-³.
To calculate the molarity of NaOH, we can use the same formula: Molarity (M) = moles of solute / liters of solution. In this case, moles of NaOH = moles of HCl used for neutralization. And since the ratio is 1:1, the moles of NaOH will also be 0.45 moles.
For the concentration in g/dm³, we can use the formula: Concentration (g/dm³) = moles of solute / volume of solution (in dm³). But we only have the volume in cm³. So, first, we'll convert cm³ to dm³ and then use the formula. With the given volume of 37.0 cm³, we divide it by 1000 to convert it to dm³.
So, to summarize - the molarity of NaOH is 0.45 mol dm-³, and the concentration in g/dm³ can be calculated by dividing the moles of NaOH by the volume of solution in dm³. I hope this solution makes you say, "NaOH my, that's some chemistry!"
1. To calculate the molarity of a solution, you need to divide the number of moles of solute (in this case, sodium chloride) 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
M = 8.75 moles / 6.22 L
M ≈ 1.41 M
Therefore, the molarity of the solution is approximately 1.41 M.
2. 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 sodium hydroxide into moles:
Moles = Mass / Molar mass
Moles = 428 g / (22.99 g/mol + 16.00 g/mol + 1.01 g/mol)
Moles ≈ 10 moles
Molarity (M) = Moles of solute / Volume of solution
M = 10 moles / 6.4 L
M ≈ 1.56 M
Therefore, the molarity of the solution is approximately 1.56 M.
3. To calculate the final volume of a solution, you need to use the molarity and moles of solute.
Given:
Moles of glucose (C6H12O6) = 4.85 moles
Desired molarity = 0.75 M
Moles = Molarity x Volume
Volume = Moles / Molarity
Volume = 4.85 moles / 0.75 M
Volume ≈ 6.47 L
Therefore, the final volume of the solution will be approximately 6.47 L.
4. To calculate the molarity of the sodium hydroxide (NaOH) solution and its concentration in g/dm3, we can use the balanced equation and the titration data.
Given:
Volume of NaOH solution = 37.0 cm3 = 0.037 dm3
Volume of HCl required for neutralization = 22.0 cm3 = 0.022 dm3
Molarity of HCl solution = 0.45 mol/dm3
From the balanced equation, the mole ratio between NaOH and HCl is 1:1. This means that 1 mole of NaOH reacts with 1 mole of HCl.
Moles of HCl = Molarity x Volume
Moles of HCl = 0.45 mol/dm3 x 0.022 dm3
Moles of HCl = 0.0099 mol
Since 1 mole of NaOH reacts with 1 mole of HCl, the moles of NaOH also equal 0.0099 mol.
Molarity of NaOH solution = Moles of NaOH / Volume of NaOH solution
Molarity of NaOH = 0.0099 mol / 0.037 dm3
Molarity of NaOH ≈ 0.27 M
Concentration of NaOH in g/dm3 = Molarity x Molar mass of NaOH
Concentration of NaOH = 0.27 mol/dm3 x (22.99 g/mol + 16.00 g/mol + 1.01 g/mol)
Concentration of NaOH ≈ 16.88 g/dm3
Therefore, the molarity of the sodium hydroxide (NaOH) solution is approximately 0.27 M, and its concentration is approximately 16.88 g/dm3.
1. To calculate the molarity of a solution, you need to use the formula:
Molarity (M) = Moles of solute / Volume of solution (in liters)
In this case, you are given that 8.75 moles of sodium chloride (NaCl) are dissolved in 6.22 liters of water. To calculate the molarity, plug the values into the formula:
Molarity = 8.75 moles / 6.22 L = 1.41 M
Therefore, the molarity of the sodium chloride solution is 1.41 M.
2. To determine the molarity of a solution, you need to use the formula:
Molarity (M) = Moles of solute / Volume of solution (in liters)
In this case, you are given that 428 grams of sodium hydroxide (NaOH) are dissolved in 6.4 liters of water. To calculate the molarity, you need to convert the grams of NaOH to moles using its molar mass (40 g/mol):
Moles of NaOH = 428 g / 40 g/mol = 10.7 moles
Now, plug the values into the molarity formula:
Molarity = 10.7 moles / 6.4 L = 1.67 M
Therefore, the molarity of the sodium hydroxide solution is 1.67 M.
3. To calculate the final volume of a solution, you need to use the formula:
Final Volume = Moles of solute / Desired Molarity
In this case, you are given that 4.85 moles of glucose (C6H12O6) are to be dissolved to make a 0.75 M solution. Plug the values into the formula:
Final Volume = 4.85 moles / 0.75 M = 6.47 L
Therefore, the final volume of the solution will be 6.47 liters.
4. To calculate the molarity of a solution, you can use the equation:
Molarity (M) = Moles of solute / Volume of solution (in liters)
In this case, you are given that 37.0 cm³ of a sodium hydroxide (NaOH) solution reacts with 22.0 cm³ of 0.45 mol/dm³ hydrochloric acid (HCl).
First, convert the volume of hydrochloric acid used to liters:
Volume of HCl = 22.0 cm³ = 22.0 x 10⁻³ L = 0.022 L
Now, use the equation to calculate the moles of hydrochloric acid:
Moles of HCl = Molarity x Volume of solution = 0.45 mol/dm³ x 0.022 L = 0.0099 moles
Since the balanced equation shows a 1:1 ratio between NaOH and HCl, the moles of sodium hydroxide (NaOH) present in the solution would also be 0.0099 moles.
Now, use the volume of sodium hydroxide solution (37.0 cm³) to calculate the molarity:
Molarity = 0.0099 moles / 0.037 L = 0.27 M
The molarity of the sodium hydroxide solution is 0.27 M.
To find the concentration of sodium hydroxide in g/dm³, you can use the molarity and the molar mass of NaOH (40.0 g/mol):
Concentration = Molarity x Molar mass = 0.27 M x 40.0 g/mol = 10.8 g/dm³
Therefore, the concentration of sodium hydroxide in the solution is 10.8 g/dm³.