Calculate the volume in milliliters of a 2.8/molL zinc nitrate solution that contains 25.0g of zinc nitrate ZnNO32. Round your answer to 2 significant digits.

To solve this problem, we need to follow these steps:

1. First, we need to find the molar mass of zinc nitrate (Zn(NO3)2). The molar mass of zinc is approximately 65.38 g/mol, nitrogen is approximately 14.01 g/mol, and oxygen is approximately 16.00 g/mol. Since there are two nitrogen atoms and six oxygen atoms in zinc nitrate, we can calculate the molar mass as follows:

   Molar mass = 65.38 g/mol (for Zn) + 2 * 14.01 g/mol (for N) + 6 * 16.00 g/mol (for O) = 189.4 g/mol

2. Next, we need to convert the mass of zinc nitrate given (25.0 g) into moles. We can do this using the molar mass we just calculated:

   Moles = 25.0 g / 189.4 g/mol = 0.132 moles

3. Now, we can find the volume of the solution. We know that the concentration of the solution is 2.8 mol/L, so we can use the formula for concentration (C = n/V) to find the volume:

   Volume = Moles / Concentration = 0.132 moles / 2.8 mol/L = 0.047 L

4. Finally, we need to convert this volume from liters to milliliters. There are 1000 milliliters in a liter, so:

   Volume = 0.047 L * 1000 = 47 mL

So, the volume of the 2.8 mol/L zinc nitrate solution that contains 25.0 g of zinc nitrate is approximately 47 mL.