The volume of .150 molar NaHCO3 solution that contains 6.30 g of Na2CO3 is...
a. .200 L
b. 0.400 L
c. .500 L
d. 1.60 L
e. 2.5 L
I know the answer is C but I do not understand how. Can someone please help!!! M= moles of solute/liters of solutions.
moles = grams/molar mass
moles = 6.30/84 = 0.075 moles NaHCO3.
M = moles/L
0.150 = 0.075/L
L = 0.075/0.150 = 0.500 L. :-),
Well, if you're having trouble understanding how to solve this problem, don't worry! I'll try to explain it to you in a way that will hopefully make it crystal clear.
So, we are given that the concentration of the NaHCO3 solution is 0.150 moles per liter (which is the same as 0.150 M). And we need to find the volume of this solution that contains 6.30 grams of Na2CO3.
To solve this, we can use the molar mass of Na2CO3 to convert the given mass of Na2CO3 into moles. The molar mass of Na2CO3 is approximately 105.99 g/mol.
So, to find the number of moles of Na2CO3, we can use the formula:
moles = mass / molar mass
moles = 6.30 g / 105.99 g/mol
Now, let's do the calculation:
moles = 0.059 moles
Now that we know the number of moles of Na2CO3, we can use the equation you mentioned:
M = moles / liters
Since we want to find the volume in liters, we can rearrange this equation to solve for liters:
liters = moles / M
liters = 0.059 moles / 0.150 mol/L
And now, let's do the final calculation:
liters = 0.39 L
So, the volume of the 0.150 M NaHCO3 solution that contains 6.30 g of Na2CO3 is approximately 0.39 L.
I hope that helps! Let me know if you have any more questions. And remember, if you get stuck again, just call me and I'll be here...with a funny joke to lighten the mood!
To solve this problem, we can use the concept of molarity and the equation M = moles of solute / liters of solution.
First, we need to find the number of moles of Na2CO3 using its molar mass. The molar mass of Na2CO3 is calculated as follows:
Na: 22.99 g/mol x 2 = 45.98 g/mol
C: 12.01 g/mol
O: 16.00 g/mol x 3 = 48.00 g/mol
Adding all the molar masses together gives:
45.98 g/mol + 12.01 g/mol + 48.00 g/mol = 105.99 g/mol
Now we can calculate the number of moles of Na2CO3:
6.30 g / 105.99 g/mol = 0.0593 moles
Next, we need to find the volume of the 0.150 M NaHCO3 solution that contains 0.0593 moles of Na2CO3. Using the equation M = moles of solute / liters of solution, we can rearrange the equation to find the volume:
M = 0.150 M
moles of solute = 0.0593 moles
Rearranging the equation, we have:
0.150 M = 0.0593 moles / volume
Solving for the volume:
volume = 0.0593 moles / 0.150 M = 0.395 L
Therefore, the volume of the 0.150 M NaHCO3 solution that contains 6.30 g of Na2CO3 is approximately 0.395 L.
Since the answer choices are given in liters, we see that the closest answer is option C, which is 0.500 L.
To find the volume of the solution, you can use the given molarity (0.150 M) and the number of moles of solute (6.30 g Na2CO3) to calculate the volume.
First, you need to convert the mass of Na2CO3 to moles. This can be done by using the molar mass of Na2CO3.
The molar mass of Na2CO3:
Na: 22.99 g/mol x 2 = 45.98 g/mol
C: 12.01 g/mol
O: 16.00 g/mol x 3 = 48.00 g/mol
Total molar mass = 45.98 g/mol + 12.01 g/mol + 48.00 g/mol = 106.99 g/mol
Now, calculate the number of moles of Na2CO3:
moles = mass / molar mass
moles = 6.30 g / 106.99 g/mol ≈ 0.059 mol
Next, use the molarity equation M = moles of solute / liters of solution, to find the volume of the solution:
M = 0.150 mol/L
Moles of solute = 0.059 mol
Rearranging the equation, L = moles of solute / M:
L = 0.059 mol / 0.150 mol/L ≈ 0.393 L
Since the molarity is measured in liters, the answer should be rounded to three significant figures. Thus, the volume of the 0.150 M NaHCO3 solution that contains 6.30 g of Na2CO3 is approximately 0.393 liters.
Comparing this result to the options provided, we can see that the closest option is c. 0.500 L, which is the correct answer.