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

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

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

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 = ×.

Where Δ is the elevation of the boiling point i.e. 2.42°C.

is the boiling point elevation constant of benzene i.e. 2.53°C.kg/mol.

is the molality of the the solute i.e. the solid.

Now the can be obtained by plugging the values in the equation.

2.42°C = 2.53°C.kg/mol ×

Or, = mol/kg

Or,   = 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.

periodic trends is an atom that gets larger as the number of electronic shells increase, the radius of atoms increases as you go down a certain group in the periodic table of elements. in general, the size of an atom will decrease as you move from left to right of a certain period.