1. What is the molar of a sodium bicarbonate solution prepared from 150 g of NaHCO3 in 150 mL solution?
2. What mass of glucose (C6H12O6) is needed to prepare a solution of 10 M, 5 L solution?
3. What volume in liters of a 4.0 M Ca(NO3)2 is needed to provide 50 g Ca(NO3)2?
I'm assuming for the first question you meant molar concentration.. if this is the case use the formula c=n/v (or concentration equals mass divided by volume) so you would first find the molar mass of sodium bicarbonate (i'm assuming you know how to do that) and use the formula n= m/M or the number of moles equals the mass of the substance divided by molar mass... for numbers two and three used similar methods.. hope this helps and feel free to ask any additional questions :)
To solve these problems, we need to use the concept of molarity. Molarity (M) is a measure of the concentration of a solute in a solution and is defined as the number of moles of solute per liter of solution.
Here's how you can solve each of the given problems:
1. What is the molar of a sodium bicarbonate solution prepared from 150 g of NaHCO3 in 150 mL solution?
Step 1: Convert the given mass of NaHCO3 to moles.
To do this, we need to determine the molar mass of NaHCO3. The molar mass of NaHCO3 is calculated by summing the atomic masses of its constituent elements (Na, H, C, O). The atomic mass of Na is 22.99 g/mol, H is 1.01 g/mol, C is 12.01 g/mol, and O is 16.00 g/mol. Thus, the molar mass of NaHCO3 is 22.99 + 1.01 + 12.01 + (3 * 16.00) = 84.01 g/mol.
Next, we can calculate the number of moles:
Number of moles = given mass (g) / molar mass
In this case, it would be:
Number of moles = 150 g / 84.01 g/mol.
Step 2: Convert the given volume to liters.
The given volume is 150 mL, but we need to convert it to liters by dividing by 1000.
150 mL / 1000 = 0.150 L.
Step 3: Calculate the molarity (M).
Molarity (M) = moles of solute / liters of solution
In this case, it would be:
Molarity (M) = (150 g / 84.01 g/mol) / 0.150 L.
2. What mass of glucose (C6H12O6) is needed to prepare a solution of 10 M, 5 L solution?
Step 1: Determine the molar mass of glucose (C6H12O6).
The molar mass of glucose (C6H12O6) can be calculated by summing the atomic masses of its constituent elements (C, H, O). The atomic mass of C is 12.01 g/mol, H is 1.01 g/mol, and O is 16.00 g/mol. Thus, the molar mass of glucose is 6 * 12.01 + 12 * 1.01 + 6 * 16.00 = 180.18 g/mol.
Step 2: Calculate the number of moles needed.
Molarity (M) = moles of solute / liters of solution
In this case, it would be:
10 M = moles / 5 L.
moles = 10 M * 5 L.
Step 3: Convert moles to grams.
Mass (g) = moles * molar mass.
In this case, it would be:
Mass (g) = 10 M * 5 L * 180.18 g/mol.
3. What volume in liters of a 4.0 M Ca(NO3)2 is needed to provide 50 g Ca(NO3)2?
Step 1: Convert the given mass of Ca(NO3)2 to moles.
To do this, we need to determine the molar mass of Ca(NO3)2. The molar mass of Ca(NO3)2 is calculated by summing the atomic masses of its constituent elements (Ca, N, O). The atomic mass of Ca is 40.08 g/mol, N is 14.01 g/mol, and O is 16.00 g/mol. Thus, the molar mass of Ca(NO3)2 is 40.08 + (2 * 14.01) + (6 * 16.00) = 164.09 g/mol.
Next, we can calculate the number of moles:
Number of moles = given mass (g) / molar mass.
In this case, it would be:
Number of moles = 50 g / 164.09 g/mol.
Step 2: Calculate the volume of the solution.
Molarity (M) = moles of solute / liters of solution.
In this case, it would be:
4.0 M = (50 g / 164.09 g/mol) / liters.
liters = (50 g / 164.09 g/mol) / 4.0 M.
I hope this helps you understand how to solve these types of problems! Let me know if you have any further questions.