To determine the percentage change in the molarity of the solution, we need to follow these steps:

Calculate the initial volume of the solution at $$+90^{\circ} \mathrm{C}$$.
Calculate the final volume of the solution at $$+10^{\circ} \mathrm{C}$$.
Determine the initial and final molarities of the solution.
Compute the percentage change in the molarity.

Given information:

Mass of the solute (substance) = 50 g
Mass of the solvent (water) = 1 kg = 1000 g
Initial density = $$1.1 \mathrm{~g} \mathrm{~cc}^{-1}$$
Final density = $$1.15 \mathrm{~g} \mathrm{~cc}^{-1}$$

Step 1: Calculate the initial volume of the solution at $$+90^{\circ} \mathrm{C}$$.

The total mass of the solution is:

$$\text{Total mass} = 50 \, g + 1000 \, g = 1050 \, g$$

Using the initial density, the initial volume is:

$$V_{\text{initial}} = \frac{\text{Total mass}}{\text{Initial density}} = \frac{1050 \, g}{1.1 \, \mathrm{g/cc}} \approx 954.545 \, \mathrm{cc}$$

Step 2: Calculate the final volume of the solution at $$+10^{\circ} \mathrm{C}$$.

Using the final density, the final volume is:

$$V_{\text{final}} = \frac{\text{Total mass}}{\text{Final density}} = \frac{1050 \, g}{1.15 \, \mathrm{g/cc}} \approx 913.043 \, \mathrm{cc}$$

Step 3: Determine the initial and final molarities of the solution.

Molarity (M) is defined as the number of moles of solute per liter of solution. Let's assume the molar mass (M_s) of the solute (substance) is required (denote it as M_s).

Number of moles of the solute:

$$\text{Number of moles} = \frac{50 \, g}{M_s \, \mathrm{g/mol}} = \frac{50}{M_s} \, \mathrm{mol}$$

Initial molarity:

$$M_{\text{initial}} = \frac{\text{Number of moles}}{\text{Initial volume in liters}} = \frac{\frac{50}{M_s}}{\frac{954.545}{1000}} = \frac{50 \times 1000}{M_s \times 954.545} \approx \frac{50 \times 1.047}{M_s}$$

Final molarity:

$$M_{\text{final}} = \frac{\text{Number of moles}}{\text{Final volume in liters}} = \frac{\frac{50}{M_s}}{\frac{913.043}{1000}} = \frac{50 \times 1000}{M_s \times 913.043} \approx \frac{50 \times 1.095}{M_s}$$

Step 4: Compute the percentage change in molarity.

Percentage change in molarity:

$$\text{Percentage change} = \frac{M_{\text{final}} - M_{\text{initial}}}{M_{\text{initial}}} \times 100$$

Substituting the values:

$$\text{Percentage change} = \frac{\frac{50 \times 1.095}{M_s} - \frac{50 \times 1.047}{M_s}}{\frac{50 \times 1.047}{M_s}} \times 100 = \frac{50 \times (1.095 - 1.047)}{50 \times 1.047} \times 100$$

$$\text{Percentage change} = \frac{(1.095 - 1.047)}{1.047} \times 100 = \frac{0.048}{1.047} \times 100 \approx 4.58\%$$

Since the closest provided option is 4.5, the correct answer is:

Option B: 4.5

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