Preparing 250 mL of 0.15 M Saline
A lab practical requires each student to prepare 250.0 mL of 0.150 M NaCl (molar mass 58.44 g/mol) for a quantitative assay with a stated tolerance of ±1.0% on concentration. The bench provides solid NaCl, a balance readable to 0.001 g, a 250.0 mL volumetric flask, a 250 mL graduated cylinder, and a 1.50 M NaCl stock solution. Student A weighs the solid and dissolves it. Student B measures 25.0 mL of the 1.50 M stock, adds it to 225.0 mL of distilled water in the graduated cylinder, and stirs. Student C finds a bottle labeled "0.9% w/v NaCl" in the supply cabinet and proposes using it undiluted to save time.
1. Calculate the mass of solid NaCl needed and the volume of 1.50 M stock needed. Then evaluate Student B's procedure.
Read the worked answer
Moles required: 0.150 mol/L × 0.2500 L = 0.03750 mol.
Mass route: 0.03750 mol × 58.44 g/mol = 2.192 g (weigh ~2.192 g on the 0.001 g balance; a ±0.001 g error is about ±0.05%, inside tolerance).
Stock route: V₁ = (0.150 M × 250.0 mL) ÷ 1.50 M = 25.0 mL of stock.
Student B's error: Adding 25.0 mL of stock to 225.0 mL of water assumes volumes are additive and final volume is exactly 250.0 mL. Dissolved solute changes the solution's volume, so the true final volume is not guaranteed to be 250.0 mL, and the concentration is uncertain by more than the ±1% tolerance allows. The correct procedure dilutes to the mark: pipette 25.0 mL of stock into the 250.0 mL volumetric flask, then add solvent until the meniscus sits on the calibration line, and invert to mix.
The distinction the exam is probing is "dilute to volume" versus "dilute by volume." They are not interchangeable when a tolerance is specified.
2. Evaluate Student C's proposal to use the 0.9% w/v bottle undiluted. Support your answer with a calculation and state whether any adjustment could make that bottle usable.
Read the worked answer
Convert the label to molarity: 0.9% w/v = 0.90 g per 100 mL = 9.0 g/L. 9.0 g/L ÷ 58.44 g/mol = 0.154 M.
Compare to target: 0.154 M versus 0.150 M is about +2.6% - outside the stated ±1.0% tolerance, so using it undiluted fails the assay specification even though both solutions are "saline." The label describes a clinical product, not an analytical standard.
Can it be salvaged? Yes, by dilution: take V₁ = (0.150 M × 250.0 mL) ÷ 0.154 M ≈ 243.5 mL of the 0.9% solution and dilute to 250.0 mL in a volumetric flask. Practical caveat: the 0.9% bag is labeled to two significant figures, so its true concentration is uncertain by roughly ±0.5%, which eats most of the ±1% budget. For a graded quantitative assay, the weighed solid or the volumetric dilution of the 1.50 M stock is the better choice; the bag is acceptable only where the tolerance is looser.
Reasoning pattern: whenever a substitute reagent is offered, convert it to the same units as the specification before deciding - "it's saline too" is not an argument.