What mass of sodium hydrogen sulphate (NaHSO4) is required to prepare 200 mL of a 0.200 mol/L solution? a) Calculate the moles of sodium hydrogen sulphate present. (n=CV) b) Calculate the mass of sodium hydrogen sulphate present. (m =nM
The Correct Answer and Explanation is:
Step a: Calculate the moles of sodium hydrogen sulfate present using the formula
n = C × V
Where:
- n = number of moles
- C = concentration (mol/L)
- V = volume (in liters)
Given:
C = 0.200 mol/L
V = 200 mL = 0.200 L
Substitute into the formula:
n = 0.200 mol/L × 0.200 L = 0.040 mol
Step b: Calculate the mass of sodium hydrogen sulfate using the formula
m = n × M
Where:
- m = mass (g)
- n = number of moles
- M = molar mass of NaHSO₄
To calculate the molar mass of NaHSO₄:
- Na = 22.99 g/mol
- H = 1.01 g/mol
- S = 32.07 g/mol
- O₄ = 16.00 × 4 = 64.00 g/mol
M = 22.99 + 1.01 + 32.07 + 64.00 = 120.07 g/mol
Now calculate the mass:
m = 0.040 mol × 120.07 g/mol = 4.80 g
Final Answer:
4.80 grams of sodium hydrogen sulfate (NaHSO₄) is required.
📘 Explanation (like in textbooks):
To prepare a solution of a known concentration, it is essential to determine the exact amount of solute (in this case, NaHSO₄) needed. The molarity (mol/L) expresses how many moles of solute are dissolved in one liter of solution. Since we are preparing only 200 mL (or 0.200 L), we multiply the desired concentration by this volume to find the moles of NaHSO₄ required. This gives 0.040 mol.
Next, to convert moles into grams—a measurable quantity—we multiply the number of moles by the molar mass of NaHSO₄. The molar mass is calculated by summing the atomic masses of each element in the compound. Multiplying 0.040 mol by 120.07 g/mol gives the required mass as 4.80 grams. This process ensures precision in preparing chemical solutions for laboratory or industrial use.
