1. Given reaction and rate law

$A + B \to C$

with rate expression

$\text{rate} = k[A]^{1/2}[B]^{1/2}$

Initial concentrations are:

$[A]_0 = [B]_0 = 1\,\text{M}$

Given:

$k = 4.6 \times 10^{-2}\,\text{s}^{-1} = 0.046\,\text{s}^{-1}$

We need the time when:

$[A] = 0.1\,\text{M}$

2. Relating $[A]$ and $[B]$ during reaction

Since the stoichiometry is $1:1$ and initially both are equal, they remain equal throughout the reaction:

$[A] = [B]$

So the rate law becomes:

$\text{rate} = k[A]^{1/2}[A]^{1/2} = k[A]$

Thus the reaction behaves like a first-order reaction in $A$ under these conditions.

3. Write differential rate equation

For disappearance of $A$:

$-\frac{d[A]}{dt} = k[A]$

Separate variables:

$\frac{d[A]}{[A]} = -k\,dt$

Integrate from $[A]_0 = 1$ to $[A] = 0.1$ and from $t=0$ to $t=t$:

$\int_{1}^{0.1} \frac{d[A]}{[A]} = -k \int_0^t dt$

$\ln(0.1) - \ln(1) = -kt$

$\ln(0.1) = -kt$

Since

$\ln(0.1) = -2.303$

we get

$t = \frac{2.303}{k}$

Substitute $k=0.046\,\text{s}^{-1}$:

$t = \frac{2.303}{0.046}$

$t \approx 50.07\,\text{s}$

Nearest integer:

$\boxed{50}$

4. Comparison with stored answer

Stored correct answer = $50$

Our derived answer = $50$

So they agree.