What is the molarity of 5.00 g of NaOH in 750.0 mL of solution?
Answer (3)
The molarity of 5.00 g of NaOH in 750.0 mL of solution is 0.1667 M.
To calculate the molarity of the NaOH solution, we use the formula:
Molarity (M) = liters of solution (V) moles of solute (n)
First, we need to calculate the number of moles of NaOH. The molar mass of NaOH is 40.00 g/mol (sodium: 22.99 g/mol + oxygen: 16.00 g/mol + hydrogen: 1.01 g/mol). Using the given mass of NaOH (5.00 g), we calculate the moles as follows:
n NaOH = molar mass of NaOH mass of NaOH
n NaOH = 40.00 g/mol 5.00 g
n NaOH = 0.125 mol
Next, we convert the volume of the solution from milliliters to liters:
V = 750.0 mL × 1000 mL 1 L
V = 0.750 L
Now we can calculate the molarity:
M = V n NaOH
M = 0.750 L 0.125 mol
M = 0.1667 M
Therefore, the molarity of the solution is 0.1667 M.
The molarity of the NaOH solution is 0.17 M .
Calculate moles of NaOH:
- Given mass of NaOH = 5.00 g
- Molar mass of NaOH (Na = 23.0 g/mol, O = 16.0 g/mol, H = 1.0 g/mol):
Molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g/mol
- Moles of NaOH:
Moles = Molar mass Mass = 40.0 g/mol 5.00 g = 0.125 mol
Calculate molarity (M):
- Molarity is defined as moles of solute per liter of solution (mol/L):
Molarity = Volume of solution in liters Moles of solute
- Volume of solution given = 750.0 mL = 0.750 L
- Molarity:
Molarity = 0.750 L 0.125 mol = 0.1667 M
Convert to appropriate significant figures:
- Given that the mass of NaOH is specified to 3 significant figures (5.00 g) and the volume to 4 significant figures (750.0 mL), the molarity should be reported to 3 significant figures.
- Therefore, rounding to three significant figures gives us:
Molarity = 0.167 M
However, typically molarity is reported to two decimal places unless otherwise specified, so rounding appropriately:
Molarity = 0.17 M
Complete Question:
What is the molarity of 5.00 g of NaOH in 750.0 mL of solution?
The molarity of 5.00 g of NaOH in 750.0 mL of solution is approximately 0.17 M. This is calculated by finding moles of NaOH, converting the volume to liters, and dividing moles by volume. The final answer is reported to three significant figures as 0.167 M, or approximately 0.17 M.
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