Elements $$\%$$ by wt $$\%$$ by wt/ At wt $$=X_t$$ $$\dfrac{X_{t}}{X_{min}}$$
$$C$$ $$80.78$$ $$80.78/12=6.73$$ $$6.75/0.3475=19$$
$$H$$ $$13.36$$ $$13.36/1=13.36$$ $$13.36/0.3475=38$$
$$O$$ $$5.56$$ $$5.56/16=0.3475$$ $$0.3475/0.3475=1$$
$$\therefore$$ Empirical formula of the compound $$=C_{19}H_{38}O$$
Empirical wt $$19\times 12+38+16=282$$
Also, $$\triangle T_{f}=K_{f}\times m$$
$$(5.5-3.37)=5.12\times \dfrac{1\ g}{M.W.\times 8.5\times 1.0028}\times 1000$$
$$(d_{benzene}=1.0028\ g/cc)$$
$$\therefore M.w=\dfrac{5.12\times 1000}{8.5\times 1.0028\times 2.13}=282$$
$$\therefore$$ So $$M.F.=(C_{19}H_{38}O)_{1}=\left(\because n=\dfrac{M.w}{E.w}=\dfrac{282}{282}=1\right)$$
$$M.F=C_{19}H_{38}O$$