A student dissolved 4.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
0.0197 g
0.0394 g
0.0542 g
0.108 g
Don't understand it. I keep getting 0.394 g as my final result.
1. (0.218639 M)(0.00400 L) = (x M)(0.275 L)
= 0.00318 M Co(NO3)2
2. 0.00318 M Co(NO3)2 x 2 mol NO3- / 1 mol Co(NO3)2 x 62 g NO3- / 1 mol NO3- = 0.394 g NO3-
.108
.108 is the answer
Well, it seems like the answer you got is 0.394 g, but the options given don't include that. Looks like someone's playing a prank on you. Don't worry, I'll help you figure it out.
Let's go through the calculations again:
1. First, we find the molarity (M) of the stock solution:
M₁ = (0.00400 mol / 0.100 L) = 0.04 M Co(NO₃)₂
2. Next, we find the number of moles of Co(NO₃)₂ in the 4.00 mL taken:
moles of Co(NO₃)₂ = M₁ x V₁ = (0.04 mol/L) x (0.00400 L) = 0.00016 mol
3. Now, we need to find the concentration of the final solution (M₂):
M₂ = (moles of Co(NO₃)₂) / (total volume in liters)
M₂ = (0.00016 mol) / (0.275 L) = 0.000582 M Co(NO₃)₂
4. Finally, we find the mass of NO₃⁻ ion in the final solution:
mass of NO₃⁻ ion = (M₂ x V₂ x Molar Mass of NO₃⁻ ion)
mass of NO₃⁻ ion = (0.000582 mol/L) x (0.275 L) x (62 g/mol)
mass of NO₃⁻ ion ≈ 0.0197 g
So, the correct answer should be 0.0197 g. Now, that's nothing to joke about!
To determine the number of grams of NO3- ions in the final solution, you can follow these steps:
Step 1: Calculate the molarity of the stock solution.
Given that 4.00 g of Co(NO3)2 is dissolved in enough water to make 100. mL of stock solution, we first need to calculate its molarity (M).
Molarity (M) = moles of solute / volume of solution in liters
Convert the mass of Co(NO3)2 to moles:
moles of Co(NO3)2 = 4.00 g / molar mass of Co(NO3)2
The molar mass of Co(NO3)2 can be calculated as follows:
Co: 58.93 g/mol
N: 14.01 g/mol (from nitrate, NO3)
O: 16.00 g/mol (3 oxygens in NO3)
Molar mass of Co(NO3)2 = (58.93 g/mol) + 2 * [(14.01 g/mol) + (16.00 g/mol * 3)]
Calculate the molarity:
M = moles of Co(NO3)2 / volume of solution in liters
= (moles of Co(NO3)2) / (100 mL / 1000 mL/L)
Step 2: Calculate the moles of NO3- ions in the final solution.
Using the balanced chemical equation of Co(NO3)2, we know that for every 1 mole of Co(NO3)2, there are 2 moles of NO3- ions.
moles of NO3- ions = Molarity of Co(NO3)2 x (2 moles of NO3- / 1 mole of Co(NO3)2)
Step 3: Convert moles to grams.
To convert moles of NO3- ions to grams, we need to multiply by the molar mass of NO3-.
grams of NO3- ions = moles of NO3- ions x molar mass of NO3-
Finally:
Plug in the values from the given problem into the formulas:
Step 1:
Molarity = (0.218639 M)(0.00400 L) / (0.275 L) ≈ 0.00318 M Co(NO3)2
Step 2:
moles of NO3- ions = 0.00318 M Co(NO3)2 x (2 mol NO3- / 1 mol Co(NO3)2)
Step 3:
grams of NO3- ions = moles of NO3- ions x molar mass of NO3-
Now, calculate directly:
grams of NO3- ions = 0.00318 M Co(NO3)2 x (2 mol NO3- / 1 mol Co(NO3)2) x (62 g NO3- / 1 mol NO3-)
grams of NO3- ions ≈ 0.00318 M Co(NO3)2 x 2 x 62 g / (1 mol Co(NO3)2 x 1 mol NO3-)
grams of NO3- ions ≈ 0.396 g
Therefore, the correct answer is approximately 0.396 g of NO3- ions in the final solution, which is closest to the given option of 0.394 g.
idek
The correct answer isn't listed. Your solution doesn't work because you have convert to M and your answer is moles/L of NO3. In addition, the 275 never enters into the problem The problem asks for grams.
Dissolve 4.00 g Co(NO3)2 in 100 mL. That gives us 4/100 = 0.04 g/mL of Co(NO3)2 or 0.08 g NO3^-/mL. The final solution was made by using 4.00 mL of that solution; therefore, the final solution contained 0.08 g/mL x 4.00 mL = 0.32 grams NO3^-.