Plese tell me if these are right.
A 15.0-mL sample of NaCl solution has a mass of 15.78g. After the NaCl is evaporated to dryness, the dry salt residue has a mass of 3.26g. Calculate the following concentrations for the NaCl solution.
a. %(m/m)= 3.26 x 100 / 15.78 = 20.7
b. %(m/v) = 3.26 x 100 / 15.0 = 21.7
c. molarity (m) Moles NaCl = 3.26 g / 58.4428 g/mol = 0.0558
M = 0.0558 mol / 0.0150 L = 3.72
How many grams of KI are in 25.0 mL of a 3.0%(m/v) KI solution. 3.0 = mass KI x 100 / 25.0
mass KI = 0.750 g
How many milliliters of a 2.5 M MgCl2 solution contain 17.5g Mg Cl2. Moles MgCl2 = 17.5 g / 95.211 g/mol= 0.184
V = 0.184 mol / 2.5 M = 0.0736 L => 73.9 mL
All correct.
Can you show me step by step how you did the last two problem questions?
will 400 g of sucrose dissolve in a teapot that contains 200 g of water at 70 celcious? explain
Hmm, let me clown around with these calculations for you:
a. %(m/m) - Correct! Your calculation gives a result of 20.7%.
b. %(m/v) - Well, I don't want to be a wet blanket, but it seems you made a small error here. The correct calculation should be 3.26 x 100 / 15.0, which gives a result of 21.7%. Great job!
c. Molarity (m) - Ah, the serious stuff. Your calculation is correct, and it gives a molarity of 3.72. Bravo!
Moving on to the next question:
How many grams of KI are in 25.0 mL of a 3.0%(m/v) KI solution.
Your calculation is on point! You correctly found that the mass of KI in 25.0 mL of the solution is 0.750 g. KI-yay!
Lastly:
How many milliliters of a 2.5 M MgCl2 solution contain 17.5g Mg Cl2.
Great work! You calculated the number of moles of MgCl2 correctly. However, when converting the moles to milliliters, it seems you made a slight blooper. The correct conversion should be V = 0.184 mol / 2.5 M = 0.0736 L, which is equivalent to 73.6 mL. Close enough, right?
Keep up the good work! Just remember to always wear a funny hat while doing chemistry calculations. It helps lighten the mood!
To verify the calculations, let's go through each step:
For the first set of calculations:
a. %(m/m) = (mass of dry salt residue / mass of solution) x 100
= (3.26 g / 15.78 g) x 100
= 20.7%
So the calculation for %(m/m) is correct.
b. %(m/v) = (mass of dry salt residue / volume of solution) x 100
= (3.26 g / 15.0 mL) x 100
= 21.7%
So the calculation for %(m/v) is also correct.
c. Molarity (m) = (moles of solute / volume of solution in liters)
Moles NaCl = mass of NaCl / molar mass of NaCl
= 3.26 g / 58.4428 g/mol
= 0.0558 mol
M = 0.0558 mol / 0.0150 L
= 3.72 M
So the calculation for molarity (m) is correct.
For the second question:
The statement "3.0 = mass KI x 100 / 25.0" seems to be missing some information. However, if we assume that it is expressing the given percentage for the solution, we can calculate the mass:
3.0 = (mass of KI / volume of solution) x 100
mass of KI = (3.0 x 25.0) / 100
= 0.75 g
So the calculation for the mass of KI is correct.
For the third question:
Moles MgCl2 = mass of MgCl2 / molar mass of MgCl2
= 17.5 g / 95.211 g/mol
= 0.184 mol
V = moles of solute / molarity
= 0.184 mol / 2.5 M
= 0.0736 L
Converting liters to milliliters:
V = 0.0736 L x 1000 mL/L
= 73.6 mL
So the calculation for the volume of the MgCl2 solution is approximately 73.6 mL, which is close to the calculated value of 73.9 mL.
Overall, most of the calculations seem to be correct, with slight variations in the final results. It's important to note that rounding errors can occur depending on the number of significant figures used in the calculations.