Question

What mass of solid that has a molar mass 89.0 g/mol should be added to 100.0 g of benzene to raise the boiling point of benzene by 2.42°C? (The boiling point elevation constant of benzene is 2.53°C•kg/mol.)

A)2.84 g

B)8.52 g

C)21.5 g

D)93.0 g

Answer 1

Here we have to get the amount of solid required to raise the temperature of 100 g of benzene 2.42°C.

To raise the temperature of 100 g of benzene B) 8.52 g of the solid is needed.

We know, Δ
`T_(b)` =
`K_(b)`×
`C_(m)`.

Where Δ
`T_(b)` is the elevation of the boiling point i.e. 2.42°C.

`K_(b)` is the boiling point elevation constant of benzene i.e. 2.53°C.kg/mol.

`C_(m)` is the molality of the the solute i.e. the solid.

Now the
`C_(m)` can be obtained by plugging the values in the equation.

2.42°C = 2.53°C.kg/mol ×
`C_(m)`

Or,
`C_(m)` =
`(2.42)/(2.53)` mol/kg

Or,
`C_(m)` = 0.956 mol/kg.

Now 1 mole of the solid is equivalent to 89.0 g.

Thus 0.956 mol is equivalent to 0.956×89 = 85.13 g.

If 1000 g or 1 Kg of benzene contains 85.13 g of the solid its molality will be 0.956.

Thus in 100 g or 0.10 kg benzene will contain 85.13×0.10 = 8.51 g of the solid.

Thus 8.51 g of the solid is needed to raise the temperature.

Answer 2

Given:

Mass of benzene (solvent) = 100.0g = 0.1 kg

Molar mass of solid (solute) = 89 g/mol

Elevation in b.pt (ΔTb) = 2.42 C

B.pt elevation constant k = 2.53 C.kg/mol

To determine:

The mass of the solid to be added

Step-by-step explanation:

The elevation in b,pt is proportional to the molality (m) of the solute ;

ΔTb = k.m

where m = molality = moles of solute/kg of solvent

In this case we have:

ΔTb = k(benzene) * moles of solid/kg of benzene

2.42 = 2.53 * moles of solid/0.1

moles of solid = 0.0956 moles

The molar mass of solid = 89 g/mol

Thus, mass of solid = 0.0956 moles * 89 g/mol = 8.508 g

Ans: Mass of solid to be added is 8.508 g