Let the degree of dissociation be $\alpha$.

Reaction: $AB_{2(g)} \rightleftharpoons A_{(g)} + 2B_{(g)}$

Initially, moles are: $n_{AB_2}=1, \quad n_A=0, \quad n_B=0$

At equilibrium: $n_{AB_2}=1-\alpha, \quad n_A=\alpha, \quad n_B=2\alpha$

Total moles at equilibrium: $n_{\text{total}}=(1-\alpha)+\alpha+2\alpha=1+2\alpha$

Use the ideal gas equation to find total moles at equilibrium: $PV=nRT$ Given: $P=1.9\,\text{atm}, \quad V=25\,\text{L}, \quad T=27^\circ C=300\,\text{K}, \quad R=0.08206$

So, $n_{\text{total}}=\frac{PV}{RT}=\frac{1.9\times 25}{0.08206\times 300}$

$n_{\text{total}}=\frac{47.5}{24.618}\approx 1.9295$

Therefore, $1+2\alpha=1.9295$ $2\alpha=0.9295$ $\alpha\approx 0.46475$

Find mole fractions: $y_{AB_2}=\frac{1-\alpha}{1+2\alpha}, \quad y_A=\frac{\alpha}{1+2\alpha}, \quad y_B=\frac{2\alpha}{1+2\alpha}$

Partial pressures: $P_{AB_2}=y_{AB_2}P, \quad P_A=y_AP, \quad P_B=y_BP$

Write the expression for $K_p$: $K_p=\frac{P_A(P_B)^2}{P_{AB_2}}$

Substitute partial pressures: $K_p=\frac{\left(\frac{\alpha}{1+2\alpha}P\right)\left(\frac{2\alpha}{1+2\alpha}P\right)^2}{\left(\frac{1-\alpha}{1+2\alpha}P\right)}$

Simplifying, $K_p=\frac{4\alpha^3P^2}{(1+2\alpha)^2(1-\alpha)}$

Substitute $\alpha\approx 0.46475$ and $P=1.9$:

First, $\alpha^3\approx (0.46475)^3\approx 0.1004$ $1+2\alpha\approx 1.9295 \Rightarrow (1+2\alpha)^2\approx 3.7230$ $1-\alpha\approx 0.53525$ $P^2=(1.9)^2=3.61$

Thus, $K_p\approx \frac{4\times 0.1004\times 3.61}{3.7230\times 0.53525}$

Numerator: $4\times 0.1004\times 3.61\approx 1.4498$

Denominator: $3.7230\times 0.53525\approx 1.9931$

Hence, $K_p\approx \frac{1.4498}{1.9931}\approx 0.727$

Since $K_p = x \times 10^{-2}$, $0.727 = 72.7\times 10^{-2}$

Therefore, $x \approx 73$

Final integer answer: $\boxed{73}$

Comparison with stored answer: The stored correct answer is 72TO75, which includes $73$. So the derived answer agrees with the stored answer range.