When 2 g of a non volatile hydrocarbon containing 94.4 percent carbon is dissolved in 100 g benzene, the vapour pressure of benzene at 20oC is lowered from 0.09954 bar to 0.09837 bar. If the molecular mass of the hydrocarbon is expressed as (120 + y). What is the value of y?

To solve this problem, we need to use the formula for relative lowering of vapour pressure which is given by:

ΔP/P = n1/n2

Where: ΔP is the change in vapour pressure, P is the initial vapour pressure, n1 is the number of moles of the solute (hydrocarbon), n2 is the number of moles of the solvent (benzene).

First, we calculate the change in vapour pressure (ΔP), which is the initial vapour pressure minus the final vapour pressure:

ΔP = 0.09954 bar - 0.09837 bar = 0.00117 bar

Next, we calculate the number of moles of the solute (n1). The molar mass of the hydrocarbon is given as (120 + y) g/mol. Since we have 2 g of the hydrocarbon, the number of moles is:

n1 = mass/molar mass = 2 g / (120 + y) g/mol

We also need to calculate the number of moles of the solvent (n2). The molar mass of benzene is 78.11 g/mol. Since we have 100 g of benzene, the number of moles is:

n2 = mass/molar mass = 100 g / 78.11 g/mol

Now we can substitute these values into the formula for relative lowering of vapour pressure:

0.00117 bar / 0.09954 bar = (2 g / (120 + y) g/mol) / (100 g / 78.11 g/mol)

Solving this equation for y gives:

y = 120 g/mol - (2 g * 78.11 g/mol * 0.09954 bar) / (100 g * 0.00117 bar) = 120 g/mol - 66.4 g/mol = 53.6 g/mol

So, the value of y is 53.6.