Question

A solution of 0.643 g of an organic compound in 50ml of benzene (density; 0.879 g/ml) lowers its freezing point from ${5.51}^{\circ }C$ to ${5.03}^{\circ }C$. If $K_f$ for benzene is 5.12K, calculate the molecular weight of the compound:

(Round-off the answer)

Answer 1

The formula for lowering of the freezing point is $\Delta T_f=K_f\cdot m$.

Here, we know $K_f=5.12$ and $△T_f=0.48$

Using the above equation we get $0.48=5.12\times m$

So, molality is 0.09375 m

Now, weight = density X volume

weight = 43.95 g (benzene)

$molality=\frac{molesofsolute}{weightofsolvent}$

$0.09375=\frac{molesofsolute}{0.04395}$

No of moles = 0.00412 mol

now by putting weight = 0.643 g

we get molecular weight = 156.06 g