Imagine two remedies with the very same concentration and the regular boiling allude к, (С/m) Solvent same...


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Imagine two options with the very same concentration and also the normal boiling suggest к, (С/m) Solvent very same boiling point, yet one has benzene as the solvent and (C) the other has carbon tetrachloride together the solvent. Determine benzene 80.1 2.53 the molal concentration, m (or b), and boiling point, T. Carbon tetrachloride 76.8 5.03 Т, т m °C

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Elevation of boiling point: 456=k6m ; my molality. Kb molal boiling suggest constant. AT6 = To-To come Boiling allude of equipment To Boling point of Solent.
For benzene: Ko = 2.53 "m, to = 80.1 °C because that .: T6-80.1 = (2-53)m --> (). Carbon fetouchloride: K625.03, still = 76.800 T6 -76.8 = 5.03m --() » T6–76.8 - 5.03 To-80.1 2.53 >> 2.53 T6 – 76.8*2.53 = 5.03T6 - 80./X5.03 - 80.1x5.03-768x2.53=5.03 To-2.53 Tb => 409.903-194.304 = 2.5 T6 >> 208.599 = 2.5 Tb 2) Toz 208.599 = 83.44€ 9.5 putting value of Th in o) 83. 404 – 80.1= 2.53m . 83.44-80.) – 1.32 m > m= - 2.53 - molality = 1.32m, boiling point, To=83.48
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The adhering to GCMS spectrum was of the product the a Grignard reaction between phenylmagnesium bromide and acetophenone in anhydrous conditions. The major peak in the chromatogram to be 1,1-diphenylethanol. Ns was may be to determine that peak and its fragmentation. However, ns cannot number out the identit...