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

Answer:

Concentration of ethanol required =  48.476 M

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:

$= ( 1 \ M \times \dfrac{160\ MHz }{450 \ MHz}) \times \dfrac{300 \ K}{2.2\ K}$

$= ( 1 \ M \times 0.3555 ) \times136.36}$

= 48.476 M