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#charge calculation

6 public questions tagged with this topic.

During electrolysis of aqueous K₂SO₄ with inert electrodes, 0.336 L of gas (STP) is collected at both electrodes. How ma

Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ , Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻ . Total moles = (0.336/22.4) = 0.015 mol (H₂:O₂ = 2:1), H₂ = 0.01 mol, O₂ = 0.005 mol. Charge = (0.01 × 2 + 0.005 × 4) × 96500 = 0.04 × 96500 = 3860 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws

In ESI-MS, a peptide of 1500 Da with +3 charge shows m/z of:

Electrospray mass spectrometry detects multiply charged ions where measured m/z equals neutral mass plus charge-carrying protons divided by number of charges. Mathematically, m/z = (M + z*1.0073)/z. For a peptide weighing 1500 Da with three added protons, total mass becomes approximately 1503 Da, divided by three charges yields about 501 m/z. This charge reduction principle enables detection of large peptides within limited analyzer range. Understanding this calculation is essential for manual deconvolution and interpreting charge-state envelopes observed in protein ESI spectra during proteomic experiments.

Ref: NCERT Biology Class XII Principles on Klenow fill-in labeling, Lehninger Chapter 9 DNA cloning techniques, and Molecular Cloning by Sambrook Chapter 10 documenting end-labeling of cohesive termini.