To solve the problem of finding the molality of a [tex]\(0\%\)[/tex] by mass [tex]\( \text{H}_2 \text{SO}_4 \)[/tex] solution with a density of [tex]\(1.07 \, \text{g/mL}\)[/tex], let's proceed step-by-step.
### Step 1: Understanding Weight Percent
The weight percent given in the problem is [tex]\(0\%\)[/tex]. This means there is no [tex]\( \text{H}_2 \text{SO}_4 \)[/tex] present in the solution – it's purely solvent (e.g., water or another solvent).
### Step 2: Definition of Molality
Molality is defined as the number of moles of solute per kilogram of solvent. The formula for molality ([tex]\(m\)[/tex]) is:
[tex]\[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \][/tex]
### Step 3: Analyzing the Problem
Since the weight percent of [tex]\( \text{H}_2 \text{SO}_4 \)[/tex] in the solution is [tex]\(0\%\)[/tex], it implies that there are zero moles of [tex]\( \text{H}_2 \text{SO}_4 \)[/tex] in the solution.
### Step 4: Calculation of Molality
Given the fact that there are zero moles of [tex]\( \text{H}_2 \text{SO}_4 \)[/tex], the molality is calculated as:
[tex]\[ m = \frac{0 \, \text{moles}}{\text{kilograms of solvent}} \][/tex]
### Conclusion
Since the numerator is zero, the whole fraction results in zero. Therefore, the molality of the solution is:
[tex]\[ \boxed{0} \][/tex]
This detailed step-by-step explanation concludes that the molality of a [tex]\(0\%\)[/tex] by mass [tex]\( \text{H}_2 \text{SO}_4 \)[/tex] solution with a density of [tex]\(1.07 \, \text{g/mL}\)[/tex] is [tex]\(0\)[/tex].
