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Dilution of a brine stock to a mass percent

SCH3U Grade 11 Chemistry · Solutions and Solubility

On the food-science bench a volumetric pipette, a 500.0 mL volumetric flask, and a bottle of concentrated sodium chloride stock stand ready on a tray; a working brine is needed for the day's salting trials. The technician rinses the pipette with the 2.50 mol/L stock, draws it up past the graduation mark, and delivers exactly 40.0 mL into the flask. Distilled water follows in stages, a swirl after each addition, until the meniscus sits on the etched line, and the stoppered flask is inverted a dozen times to mix. A density check on the finished solution reads 1.005 g/mL, so the flask holds 502.5 g of solution. Taking NaCl as 58.44 g/mol, find the concentration after dilution, the moles of salt in the flask, the mass of that salt, and the concentration expressed as a percent by mass.

2.50 mol/L stock40.0 mL500.0 mL mark0.200 mol/L1.16% NaCl by mass

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • C₁ = 2.5 mol/L — Sodium chloride stock
  • V₁ = 40 mL — Stock pipetted in
  • V₂ = 500 mL — Flask made up to the mark
  • m_soln = 502.5 g — Mass of the diluted solution (at 1.005 g/mL)
  • M = 58.44 g/mol — Molar mass of NaCl
Determine
  1. (a)the concentration after dilution
  2. (b)the moles of salt in the flask
  3. (c)the mass of that salt
  4. (d)the concentration as a percent by mass
Step 1 of 4(a) · solve for Final concentration

Adding water changes nothing about how much salt is present — only how far it is spread. C₁V₁ = C₂V₂ is that sentence written as algebra, and it is the only step that can come first.

C1V1C2V2
Rearranged for C2
C2=C1V1V2C_2 = \frac{C_1 V_1}{V_2}
Your values, in your units
C2=(2.5 M) (40 mL)(500 mL)C_2 = \frac{\left( 2.5\ \text{M} \right) \, \left( 40\ \text{mL} \right)}{\left( 500\ \text{mL} \right)}
Converted to base units
C2=(2.5 M) (0.04 L)(0.5 L)C_2 = \frac{\left( 2.5\ \text{M} \right) \, \left( 0.04\ \text{L} \right)}{\left( 0.5\ \text{L} \right)}
Answer
C2=200 mol/m3C_2 = 200\ \text{mol/m}^{3}

Carried onward at full precision, not this rounded figure.

Open the Dilution Equation (C1V1 = C2V2) solver →

Step 2 of 4(b) · solve for Amount of solute

Multiply the diluted concentration by the full 500.0 mL. A useful check: the same moles come out of 2.50 mol/L × 40.0 mL, because the dilution never created or destroyed any solute.

VCn
Rearranged for n
n=C Vn = C \, V
200 mol/m³carried from step 1
Your values, in your units
n=(200 mol/m3) (500 mL)n = \left( 200\ \text{mol/m}^{3} \right) \, \left( 500\ \text{mL} \right)
Converted to base units
n=(0.2 M) (0.5 L)n = \left( 0.2\ \text{M} \right) \, \left( 0.5\ \text{L} \right)
Answer
n=100 mmoln = 100\ \text{mmol}

Carried onward at full precision, not this rounded figure.

Open the Molarity (C = n/V) solver →

Step 3 of 4(c) · solve for Mass

Moles to grams. This is the number a recipe, a label, or a purchasing order can actually use.

mMn
Rearranged for m
m=n Mm = n \, M
100 mmolcarried from step 2
Your values, in your units
m=(0.1 mol) (58.44 g/mol)m = \left( 0.1\ \text{mol} \right) \, \left( 58.44\ \text{g/mol} \right)
Answer
m=5.844 gm = 5.844\ \text{g}

Carried onward at full precision, not this rounded figure.

Open the Moles from Mass (n = m/M) solver →

Step 4 of 4(d) · solve for Mass percent

Food and industrial labels quote percent by mass, not molarity. Divide by the mass of the whole solution — 502.5 g, salt included — never by the mass of the water alone.

msolutemsolutionc
Rearranged for c
c=msolutemsolution×100%c = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 100\%
5.844 gcarried from step 3
Your values, in your units
c=(0.005844 kg)(502.5 g)×100%c = \frac{\left( 0.005844\ \text{kg} \right)}{\left( 502.5\ \text{g} \right)} \times 100\%
Converted to base units
c=(0.005844 kg)(0.5025 kg)×100%c = \frac{\left( 0.005844\ \text{kg} \right)}{\left( 0.5025\ \text{kg} \right)} \times 100\%
Answer
c=1.163 %c = 1.163\ \text{\%}

Carried onward at full precision, not this rounded figure.

Open the Mass Percent of a Solution solver →

Answer

Therefore the working brine stands at 0.200 mol/L, the flask holds 0.100 mol of salt — 5.844 g of it — and against the 502.5 g of solution that reads 1.16% by mass on the label.

Why this order

Dilution is the one calculation in a solutions unit that needs no chemistry at all, only bookkeeping: the solute is conserved, so concentration times volume is the same before and after. That is why step 1 leads. Everything after it is a change of currency — moles, then grams, then percent — describing one unchanging quantity of sodium chloride four different ways. Students who compute the moles from the stock concentration and the final volume get an answer 12.5 times too large, and the mistake survives all the way to the label.

The last step is where the unit systems collide. Molarity counts particles per volume and changes with temperature as the glass and the liquid expand; mass percent counts grams per gram and does not. Crossing between them requires the density, which is why the scenario has to supply it — you cannot get from 0.200 mol/L to 1.16% by mass without knowing what a litre of this particular brine weighs. That is also the reason concentrated acids ship with both figures printed on the bottle. And note what the site does with a percent: the concentration unit type is canonically the percent itself, so 1.16% carries as 1.16, not 0.0116.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.