Bond distance in $$HF$$  is $$9.17 \times {10^{ - 11}}m.$$    Dipole moment of $$HF$$  is $$6.104 \times {10^{ - 30}}Cm.$$    The percentage ionic character in $$HF$$  will be :
( electron charge $$ = 1.60 \times {10^{ - 19}}C$$    )

Solution :
$$\eqalign{ & {\text{Given}}\,\,e = 1.60 \times {10^{ - 19}}C \cr & d = 9.17 \times {10^{ - 11}}m \cr & {\text{From}}\,\,\mu = e \times d \cr & \mu = 1.60 \times {10^{ - 19}} \times 9.17 \times {10^{ - 11}} \cr & = 14.672 \times {10^{ - 30}} \cr & \% {\text{ ionic character}} \cr & {\text{ = }}\frac{{{\text{Observed dipole moment}}}}{{{\text{Calculated Dipole moment}}}} \times 100 \cr & = \frac{{6.104 \times {{10}^{ - 30}}}}{{14.672 \times {{10}^{ - 30}}}} \times 100 \cr & = 41.5\% \cr} $$