I was just wanting to make sure I did these right.
How many grams of solute are in each of the following solutions?
a.) 2.5 L of 3.0 M Al(NO3)3 solution.
for this one I get 1600 g of Al(NO3)3
b.) 75 ml of 0.5 M C6H12O6 solution.
and for this one I get 6.8g of C6 H12 O6
How many milliliters of a 12% (v/v) propyl alcohol solution would you need to obtain 4.5 ml of propyl alcohol.
I get 38 ml of solution
A student prepared a solution by dissolving 5 g of KI in enough water to give a final volume of 250 ml. what is the mass/volume percent (m/v) of KI solution.
I got 2% (m/v)KI
For #1 I obtained 1597 g and I understand why you rounded to 1600 g (because you are allowed only two significant figures); however, the 1600, leaves the reader (or instructor) in a quandary. He/she doesn't know if the last two zeros are significant or not. Technically it means they are not and if that is what you meant, (and your instructor knows that, too) great. A better way to write the answer is 1.6E2 and that leaves no doubt how many s.f. there are.
The others look ok.
Wow, you really did your math homework! Good job! It looks like you nailed those calculations. If I had a gold star sticker, I'd put it right here on your paper. Keep up the good work!
a.) To find the number of grams of solute in a solution, multiply the volume of the solution (in liters) by the molar concentration (M) and the molar mass (g/mol) of the solute.
Given:
Volume of solution (V) = 2.5 L
Molar concentration (M) = 3.0 M
Molar mass of Al(NO3)3 = 213.0 g/mol
Calculation:
Grams of Al(NO3)3 = Volume (L) × Molar concentration (M) × Molar mass (g/mol)
Grams of Al(NO3)3 = 2.5 L × 3.0 M × 213.0 g/mol
Grams of Al(NO3)3 = 1603.5 g
b.) Similar to the previous question, to find the grams of solute in a solution, multiply the volume of the solution (in liters) by the molar concentration (M) and the molar mass (g/mol) of the solute.
Given:
Volume of solution (V) = 75 mL (convert to liters: 75 mL ÷ 1000 mL/L = 0.075 L)
Molar concentration (M) = 0.5 M
Molar mass of C6H12O6 = 180.2 g/mol
Calculation:
Grams of C6H12O6 = Volume (L) × Molar concentration (M) × Molar mass (g/mol)
Grams of C6H12O6 = 0.075 L × 0.5 M × 180.2 g/mol
Grams of C6H12O6 = 6.76 g (rounded to 2 decimal places)
c.) To find the volume of a 12% (v/v) solution needed to obtain a certain volume of propyl alcohol, set up a proportional equation.
Given:
Volume of propyl alcohol needed = 4.5 mL
Percentage of propyl alcohol in the solution = 12% (v/v)
Let x be the volume of the 12% solution needed.
Proportional equation: 12/100 = 4.5/x
Calculation:
Cross-multiply: 12x = 100 × 4.5
12x = 450
x = 37.5 mL (rounded to the nearest mL)
Therefore, you would need approximately 37.5 mL of the 12% propyl alcohol solution to obtain 4.5 mL of propyl alcohol.
d.) To find the mass/volume percent (m/v) of a solution, divide the mass of solute (in grams) by the volume of the solution (in milliliters) and multiply by 100.
Given:
Mass of KI = 5 g
Volume of solution = 250 mL
Calculation:
Mass/volume percent (m/v) = (Mass of solute / Volume of solution) × 100
Mass/volume percent (m/v) = (5 g / 250 mL) × 100
Mass/volume percent (m/v) = 2% (rounded to the nearest whole number)
Therefore, the mass/volume percent (m/v) of the KI solution is approximately 2%.
To ensure accuracy and make sure you did the calculations correctly, let's go through each problem step by step and calculate the solutions together.
a.) 2.5 L of 3.0 M Al(NO3)3 solution.
To find the number of grams of solute in the solution, you need to use the molar mass of Al(NO3)3. The molar mass of Al(NO3)3 is 213 g/mol.
First, find the number of moles of Al(NO3)3 in the solution:
2.5 L * 3.0 mol/L = 7.5 mol
Next, calculate the mass of Al(NO3)3:
7.5 mol * 213 g/mol = 1597.5 g
Rounding to the appropriate number of significant figures, the answer is approximately 1600 g of Al(NO3)3.
b.) 75 ml of 0.5 M C6H12O6 solution.
Similarly, to find the number of grams of solute in the solution, you need to use the molar mass of C6H12O6. The molar mass of C6H12O6 is 180 g/mol.
First, find the number of moles of C6H12O6 in the solution:
75 ml * 0.5 mol/L = 37.5 mmol (millimoles)
Next, calculate the mass of C6H12O6:
37.5 mmol * (1/1000) mol/mmol * 180 g/mol = 6.75 g
Rounding to the appropriate number of significant figures, the answer is approximately 6.8 g of C6H12O6.
c.) How many milliliters of a 12% (v/v) propyl alcohol solution would you need to obtain 4.5 ml of propyl alcohol.
To calculate this, you can set up a proportion using the percent volume/volume (% v/v) equation:
% v/v = (volume of solute / volume of solution) * 100
Let's solve for the volume of solution (V_solution):
12% v/v = (4.5 ml propyl alcohol / V_solution) * 100
Rearranging the equation to solve for V_solution:
V_solution = (4.5 ml propyl alcohol) / (12% v/v / 100)
V_solution = (4.5 ml propyl alcohol) / 0.12
V_solution ≈ 37.5 ml
To obtain 4.5 ml of propyl alcohol, you would need approximately 37.5 ml of a 12% propyl alcohol solution.
d.) A student prepared a solution by dissolving 5 g of KI in enough water to give a final volume of 250 ml. What is the mass/volume percent (m/v) of KI solution?
To calculate the mass/volume percent (m/v) of KI solution, use the following formula:
% m/v = (mass of solute / volume of solution) * 100
Given:
Mass of KI = 5 g
Volume of solution = 250 ml
Plugging in the values into the formula:
% m/v = (5 g / 250 ml) * 100
% m/v = 2%
The mass/volume percent (m/v) of the KI solution is 2%.
I hope this helps clarify and verify your calculations. If you have any further questions, feel free to ask!