BGPT: A solution of valine contains 2.94 g/L. Calculate its molarity in mmol/L

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Convert using molar mass: molarity (mol/L) = 2.94 g/L ÷ 117.15 g/mol = 0.0251 mol/L = **25.1 mmol/L**.

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Valine molarity from g/L

Given: 2.94 g/L valine. Find molarity in mmol/L.

Step-by-step calculation

Valine molar mass (from standard atomic composition) ≈ 117.15 g/mol.
Molarity in mol/L:

M = $\frac{2.94\ \text{g/L}}{117.15\ \text{g/mol}}$ = 0.0251 $\text{mol/L}$
Convert to mmol/L:

0.0251 mol/L × 1000 = 25.1 mmol/L

Answer: 25.1 mmol/L

Sanity check

2.94 g/L is about 1/40 of 117.15 g/mol, so you expect ~0.025 mol/L → ~25 mmol/L. That matches.
No dilution factor or solution density is given, so we treat 2.94 g/L as already-referenced mass concentration per liter.

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Updated: April 11, 2026

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Key Insight

Turning mass concentration (g/L) into molar concentration (mmol/L) is exactly what you need to compare biochemical conditions across experiments that report different units.

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Compute molarity and mmol/L from g/L using valine molar mass, then format results with significant figures and unit tests for dimensional consistency.

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Assuming g/L can always be converted to mmol/L without any ambiguity about what “L” refers to (e.g., solution volume vs. solvent volume) is only safe when the problem explicitly states concentration per liter of solution.

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Prepare multiple valine solutions with the same nominal g/L, then experimentally verify density and measure whether the true molarity deviates from the theoretical mmol/L by more than analytical uncertainty.

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Discussion

BGPT Bias

I default to ideal-solution unit conversions and may under-emphasize how “per liter” definitions (solution vs. solvent) can matter if not stated.

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