1. A chemist prepares a solution by dissolving
4.603 g of NaNO3 in enough water to make
200 mL of solution. What molar concentration of sodium nitrate should appear on the
label?
Answer in units of M
2.If the chemist mistakenly makes 225 mL of
solution instead of the 200 mL, what molar concentration of sodium nitrate will the
chemist have actually prepared?
Answer in units of M
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To calculate the molar concentration of a solution, we need to use the equation:
Molarity (M) = moles of solute / volume of solution in liters
First, let's solve the first question:
1. Moles of NaNO3:
To find the moles of NaNO3, we need to divide the given mass by its molar mass. The molar mass of NaNO3 can be calculated by summing the atomic masses of sodium (Na), nitrogen (N), and three oxygen (O) atoms:
Na = 22.99 g/mol, N = 14.01 g/mol, O = 16.00 g/mol.
Molar mass of NaNO3 = (22.99 g/mol) + (14.01 g/mol) + (3 * 16.00 g/mol) = 85.00 g/mol
Moles of NaNO3 = 4.603 g / 85.00 g/mol = 0.0541 mol
2. Volume of solution in liters:
The given volume is 200 mL, but we need to convert it to liters by dividing it by 1000 mL/L.
Volume of solution = 200 mL / 1000 mL/L = 0.200 L
Now we can calculate the molar concentration:
Molarity (M) = moles of solute / volume of solution
Molarity (M) = 0.0541 mol / 0.200 L = 0.2705 M
Therefore, the molar concentration of sodium nitrate that should appear on the label is 0.2705 M.
Now let's solve the second question:
1. Volume of solution in liters:
The given volume is 225 mL, and we need to convert it to liters by dividing it by 1000 mL/L.
Volume of solution = 225 mL / 1000 mL/L = 0.225 L
2. Moles of NaNO3:
We already know the number of moles of NaNO3 is 0.0541 mol.
Now we can calculate the molar concentration:
Molarity (M) = moles of solute / volume of solution
Molarity (M) = 0.0541 mol / 0.225 L = 0.2404 M
Therefore, if the chemist mistakenly makes 225 mL of solution instead of 200 mL, the molar concentration of sodium nitrate prepared will actually be 0.2404 M.