question archive The standard for a HPLC investigation was prepared as follows: A weight of 192
The standard for a HPLC investigation was prepared as follows: A weight of 192
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The standard for a HPLC investigation was prepared as follows: A weight of 192.6 mg of pure reference material (Wstd) was mixed with 2.0 mL ("V is std") of an internal standard stock solution and the volume was then made up to 25 mL ("V1"). This intermediate standard solution was subjected to two more dilution steps: 5mL ("V2") of the first intermediate standard solution were diluted to 25 mL ("V3") and then 4 mL ("V4") of the latter solution were diluted to 25 mL ("V5"), to yield the standard solution that was injected to HPLC.
The first part of the investigation required an External Standard analysis. Identify which standard concentration is employed in the External Standard analysis and report its value. Enter your answer in regular (decimal, not scientific) format, with 4 (four) significant figures and no units.
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When 192.6 mg of standard was initially dissolved in 2 ml of internal standard stock solution and then finally the volume was made upto 25 ml, then it signifies that 25 ml (V1) of intermediate standard solution contains 192.6 mg of standard.
When 5 ml (V2) of this intermediate standard solution will contain = (192.6 mg x 5 ml)/ 25 ml
= 38.52 mg of standard
This 5 ml of intermediate was diluted and the volume was made upto 25 ml (V3). So the concentration of this dilution is 38.52 mg of standard present.
The 4 ml (V4) of the above dilution (V3) will contain = (38.52 mg x 4 ml)/ 25 ml
= 6.1632 mg of standard
So, when 4 ml (V4) of the above diluted solution (V3) when diluted to 25 ml, the new diluted solution V5 will contain 6.1632 of the standard.
