Question

If along $\mathbf{15}\mathbf{cm}$HPCL column has a place height of 5.0 what will be the half-width (in seconds) of a peak eluted at 10.0min? if plate height$\mathbf{5}\mathbf{\mu m}$,what will be${\mathbf{w}}_{\mathbf{1}\mathbf{/}\mathbf{2}}$?

Short answer

The half-width (in seconds) of a peak eluted.

$$\begin{gathered} \mathrm{L}=15\mathrm{cm} \\ \mathrm{H}=5.0\mathrm{\mu m}=5.0.{10}^{-\mathrm{t}}\mathrm{cm} \\ =\frac{15\mathrm{cm}}{5.0.{10}^{-4}\mathrm{cm}} \\ =3.{10}^4 \\ 3.{10}^4\mathrm{and}2.{05}^{-5}\mathrm{s} \end{gathered}$$

Step-by-step solution

definition of Theoretical plate

Theoretical plate height is a measure of a column's separating efficiency, similar to the number of separating plates in a liquid distillation column.

Given,

$$\begin{gathered} \mathrm{L}=15\mathrm{cm} \\ \mathrm{H}=5.0\mathrm{\mu m}=5.0.{10}^{-\mathrm{t}}\mathrm{cm} \end{gathered}$$

First we find the number of theoretical plates, N:

$$\begin{gathered} \mathrm{H}=\frac{\mathrm{L}}{\mathrm{N}}\Rightarrow \mathrm{N}=\frac{\mathrm{L}}{\mathrm{H}} \\ \mathrm{N}=\frac{\mathrm{L}}{\mathrm{N}} \\ =\frac{15\mathrm{cm}}{5.0.{10}^{-4}\mathrm{cm}} \\ =3.{10}^4 \\ {\mathrm{t}}_{\mathrm{r}}=10.0\min =600\mathrm{s} \end{gathered}$$

Divide the value

Use this and the equation below to find and calculate the half-width in seconds:

$$\begin{gathered} \mathrm{N}=5.55{\left(\frac{{\mathrm{T}}_{\mathrm{r}}}{{\mathrm{w}}_{1/2}}\right)}^2 \\ {\mathrm{w}}_{1/2}=\frac{5.{55}^2-}{{\mathrm{N}}^2}{\mathrm{t}}_{\mathrm{r}} \\ =2.05.{10}^{-5}\mathrm{s} \\ \mathrm{Hence}, \\ 3.{10}^4\mathrm{and}2.05.{10}^{-5}\mathrm{s} \end{gathered}$$