Final answer:
To determine the volume of a 1.48 M zinc nitrate solution that contains 175 g of zinc nitrate, we convert mass to moles, divide by molarity, and convert liters to milliliters. The result is 624 mL.
Step-by-step explanation:
To calculate the volume in millimeters of a 1.48m zinc nitrate solution that contains 175 g of zinc nitrate (Zn(NO3)2), we must first determine the concentration of the zinc nitrate solution in terms of molarity. Molarity (M) is defined as moles of solute per liter of solution. The molar mass of zinc nitrate is 189.41 g/mol.
First, we convert the mass of zinc nitrate to moles:
175 g Zn(NO3)2 × (1 mol / 189.41 g) = 0.924 moles Zn(NO3)2
Then, we use the molarity of the solution to find its volume:
(0.924 moles Zn(NO3)2) / (1.48 M) = 0.624 L
Finally, we convert liters to milliliters to get our final answer:
0.624 L × (1000 mL / 1 L) = 624 mL
Thus, the volume of the 1.48 M zinc nitrate solution needed to obtain 175 g of zinc nitrate is 624 mL.