Introduction

Calculating the concentration of an oligonucleotide from given moles and volume is a core skill in molecular biology and biochemistry, especially when preparing working stocks for PCR and cloning.^{[web:11][web:24]}

This example, 20 nanomoles oligonucleotide in 200 microliters concentration, illustrates how to move between units and apply the molarity formula correctly for everyday lab calculations.^{[web:23][web:25]}

Step‑by‑Step Solution

The Question

“A lyophilized sample of 20 nanomoles of an oligonucleotide is dissolved in water and the volume of the solution is made up to 200 µL. The concentration (in µM) of the oligonucleotide in this solution is _____ (in integer).”

The concentration of the oligonucleotide solution is 100 µM.

1. Recall the molarity formula

Molar concentration (molarity) is defined as:
C=n/v
where C is concentration in mol/L, $n$ is amount of substance in moles, and V is volume of solution in liters.

2. Convert units to moles and liters

Amount of substance:

• 20 nanomoles = 20 nmol.
• 1 nmol = 10^-9 mol, so 20 nmol = 20 × 10^-9 mol = 2.0 × 10^-8 mol.

Volume of solution:

• Given volume = 200 µL.
• 1 µL = 10^-6 L, so 200 µL = 200 × 10^-6 L = 2.0 × 10^-4 L.

3. Apply the formula in mol/L

C = 2.0 × 10^-8 mol/2.0 × 10^-4 L= 1.0 × 10^-4 mol/L.

So the concentration is 1.0 × 10^-4 mol/L.

4. Convert mol/L to µM

1 mol/L is equal to 10⁶ micromolar (µM).

Therefore, 1.0 × 10^-4 mol/L = 1.0 × 10^-4 × 10⁶ µM = 1.0 × 10² µM = 100 µM.

Hence, the required concentration in integer form is: 100 µM.

Concept Recap for SEO and Learning

The key idea is that molarity is always equal to moles of solute divided by volume of solution in liters, so every concentration problem reduces to unit conversion plus this simple ratio.

A typical workflow is to convert nanomoles to moles, convert microliters to liters, use ( C = n/V ) to obtain mol/L, and then convert mol/L to the required unit such as micromolar (µM).

For quick checks, nanomoles per microliter is numerically equal to millimolar, and scaling the volume shows that 20 nmol in 200 µL reasonably corresponds to a moderately concentrated stock of 100 µM for routine experiments.