Question

a) Formaldehyde (CH2=O) is the simplest aldehyde and is miscible in ethanol. i) Would the following...

a) Formaldehyde (CH2=O) is the simplest aldehyde and is miscible in ethanol.

i) Would the following intermolecular interactions be present or absent for an ethanol-formaldehyde mixture: hydrogen bonding, London dispersion, ion-ion, and dipole-induced dipole. [8 marks]

ii) 66 g of formaldehyde is dissolved in 800 g of ethanol. Given that the density of ethanol 789 g L-1, calculate the molarity, molality, and mole fraction for this solution. [20 marks]

iii) Ethanol has a freezing point depression coefficient Kfp = -1.99 K kg mol-1. Given that the freezing point of pure ethanol is -114.6°C, what temperature does this ethanol-formaldehyde mixture freeze at? [22 marks]

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Answer #1

Formaldehyde is miscible in ethanol

H​​​​​​2​​​​​C=O- - - - HO-CH2-CH3

The interactions are present in molecule such as hydrogen bonding, London Forces, ion ion interactions and induced dipole interactions. c=O bond of formaldehyde exists as a dipole. Since, oxygen is more electronegative than carbon it withdraw electron pair towards itself. It becomes partially negatively charged and carbon become partially positively charged.

2). MOLARITY:

GIVEN mass of formaldehyde= 66g

Molar mass of formaldehyde=30g / mol

Number of moles of formaldehyde= 66/30=2.2 moles

Mass of solution=66+800=866g

Density of ethanol=789g/L

Volume of 800g of ethanol= 800/789~1Li.e 1000mL

Molarity= no. Of moles/ volume of solution in L=2.2M

MOLALITY:

NO. OF moles= 2.2

Mass of solvent=800g

Molality=2.2×1000/800mol/g=2.75m

MOLE FRACTION:

Mole fraction of formaldehyde:

No. Of moles of formaldehyde=2.2moles

Molar mass of ethanol=46g/mol

No. Of moles of ethanol=800/46=17.39 moles

Mole fraction of A=no. Of moles of A/ total no. Of moles(A+B)

Mole fraction of formaldehyde=2.2/2.2+17.39)=2.2/19.59=0.11

Total mole fraction=1

Mole fraction of ethanol=1-0.11=0.89

3).∆Tf=Kf.m

∆Tf= depression in freezing point=To-Ts

Kf=constant

m= molality of solution

∆Tf=-1.99×2.75=-5.47

To=-114.6

Ts=To-∆Tf=-114.6+5.47=-109.13°C

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