Question

In: Chemistry

which of the following equations best represents methanol being dissolved in water A. CH3OH(I) —> CH3+(aq)...

which of the following equations best represents methanol being dissolved in water
A. CH3OH(I) —> CH3+(aq) + OH-(aq)
B. CH3OH(I) —> CH3O-(aq) + H+(aq)
C. CH3OH(I) —> CH3OH(aq)

What is the boiling point of a solution of 9.04 g of I2 in 75.5g of benzene
A. 78.9 C
B. 80.1 C
C. 81.3 C

Solutions

Expert Solution

When methanol is dissolved in water, water acts as a base.

This water splits methanol into its respective ions. The OH- from water reacts with H+ of methanol making another water molecule, leaving the OCH3- in its aqueous phase.

The reaction equation is:

CH3OH(l) ------> CH3O-(aq) + H+(aq).

Therefore, equation (2) is the correct answer.

02. Given,

Mass of iodine= 9.04 g

Mass of benzene= 75.5 g

Molar mass of iodine(I2) = 253.8 g/mol

Therefore,

Moles of iodine= 9.04/253.8 g = 0.0356 mol.

Now, Molality= no. Of moles of solute(iodine)/weight of solvent (in kg).

Molality (m) = 0.0356/ 0.755

m= 0.472 mol/kg.

We know the equation, ∆Tb= Kb×m

∆Tb= 0.472× 2.53

= 0.223

Therefore, TB= 80.1(original temperature)+ elevation in boiling point.

TB= 80.1+ 0.223

= 81.3 °C

Kindly comment for any doubt.

And do rate


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