Question

(20 points) i) An absorption intensity of 1.00 (in arbitrary units) is observed for the maximum peak of the 1:2:1 triplet of the 1H resonance of the CH3 group of ethanol dissolved in CCl4 at 1.00 M concentration using a 160 MHz NMR instrument at 300 K. What concentration of ethanol would you need to measure the same signal size if the measurement is performed with a 450 MHz NMR instrument at 2.2 K

Answer 1

Concentration of ethanol required = 48.476 M

Step-by-step explanation:

Given that:

the absorption intensity = 1.00

Molarity of ethanol = 1M

NMR instrument used = 160 MHz

Temperature used = 300 K

The required concentration of ethanol can be determined as follows:

`= ( 1 \ M * (160\ MHz )/(450 \ MHz)) * (300 \ K)/(2.2\ K)`

`= ( 1 \ M * 0.3555 ) *136.36}`

= 48.476 M