What is the molarity of a solution with 3.5mol of CaCl2 in 1.89L of solution
What is the molarity of a solution with 16g of CH3OH in 487mL of solution
1. M = # moles/liter of solution.
2. 16 g CH3OH/molar mass = moles.
M = moles/L.
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To find the molarity of a solution, you need to divide the number of moles of solute by the volume of the solution in liters.
1. For the first question, you have 3.5 mol of CaCl2 and 1.89 L of solution.
The molarity (M) is calculated using the formula: M = moles/volume.
M = 3.5 mol / 1.89 L = 1.85 M
Therefore, the molarity of the solution is 1.85 M.
2. For the second question, you have 16 g of CH3OH and 487 mL of solution. First, you need to convert the mass of CH3OH to moles using its molar mass.
The molar mass of CH3OH = 12.01 g/mol (C) + 1.01 g/mol (H) + 3 * 16.00 g/mol (O) = 32.04 g/mol.
Moles of CH3OH = mass / molar mass = 16 g / 32.04 g/mol = 0.499 mol.
Next, convert the volume from milliliters to liters:
487 mL = 487 mL * 1 L / 1000 mL = 0.487 L.
Now use the formula M = moles/volume to find the molarity.
M = 0.499 mol / 0.487 L = 1.02 M.
Therefore, the molarity of the solution is 1.02 M.
To find the molarity of a solution, you need to use the formula:
Molarity (M) = moles of solute / volume of solution in liters
Let's calculate the molarity for each problem.
Problem 1:
Given:
moles of CaCl2 = 3.5 mol
volume of solution = 1.89 L
Using the formula:
Molarity (M) = 3.5 mol / 1.89 L
Therefore, the molarity of the CaCl2 solution is approximately 1.85 M.
Problem 2:
Given:
mass of CH3OH = 16 g
volume of solution = 487 mL
To calculate the moles of CH3OH, we need to know its molar mass. The molar mass of CH3OH is:
12.01 g (carbon) + 3(1.01 g) (hydrogen) + 16.00 g (oxygen) = 32.04 g/mol
Now, convert the volume from milliliters (mL) to liters (L). Since 1 L = 1000 mL:
volume of solution = 487 mL / 1000 mL/L = 0.487 L
Next, use the formula:
Molarity (M) = moles of solute / volume of solution in liters
To find the moles of CH3OH:
moles of CH3OH = mass of CH3OH / molar mass of CH3OH
moles of CH3OH = 16 g / 32.04 g/mol
Now, substitute the values into the formula:
Molarity (M) = (16 g / 32.04 g/mol) / 0.487 L
Therefore, the molarity of the CH3OH solution is approximately 1.04 M.