Answer: 2
No. of moles of $HCl = \frac{0.192}{36.5}$ = 0.0053
Volume = 0.5 litre
Molarity of $HCL = \frac{0.0053}{0.5} = 1.06 \times 10^{-2}$ M
$[H^+]$ in $HCL$ solution = $1.06 \times 10^{-2} \times \frac{95}{100} = 100.7 \times 10^{-4} \approx 1 \times 10^{-2} $
Thus, $\rho H = - log [H^+] = - log (1 \times 10^{-2}) = 2$