Answer :
To calculate the volume of a zinc nitrate solution in milliliters that contains 25.0 g of zinc nitrate with a molarity of 1.9 mol/L, we follow these steps:
1. Calculate the molar mass of zinc nitrate (Zn(NO₃)₂):
- Molar mass of Zinc (Zn) = 65.38 g/mol
- Molar mass of Nitrogen (N) = 14.01 g/mol
- Molar mass of Oxygen (O) = 16.00 g/mol
The molar mass of zinc nitrate is calculated as follows:
Zn(NO₃)₂ = 1(Zn) + 2[1(N) + 3(O)]
= 1(65.38) + 2[1(14.01) + 3(16.00)]
= 65.38 + 2(14.01 + 3(16.00))
= 65.38 + 2(14.01 + 48.00)
= 65.38 + 2(62.01)
= 65.38 + 124.02
= 189.40 g/mol
2. Calculate the number of moles of zinc nitrate using the mass of zinc nitrate and its molar mass:
Moles = mass / molar mass
Moles of zinc nitrate = 25.0 g / 189.40 g/mol
Now we'll perform the division to find the number of moles:
Moles of zinc nitrate ≈ 25.0 / 189.40
≈ 0.1319 mol (rounded to four significant digits)
3. Calculate the volume of solution using the molarity and the number of moles of zinc nitrate:
Molarity (M) = moles of solute / volume of solution in liters (L)
Or equivalently,
Volume of solution (L) = moles of solute / molarity (M)
Therefore:
Volume (L) = Number of moles / Molarity
≈ 0.1319 mol / 1.9 mol/L
≈ 0.0694 L (rounded to four significant digits)
4. Convert the volume from liters to milliliters:
1 Liter = 1000 milliliters
Volume (mL) = Volume (L) * 1000
≈ 0.0694 L * 1000
≈ 69.4 mL
5. Round the answer to two significant digits:
Since the molarity was given to two significant digits, the volume should also be reported to two significant digits:
Volume (mL) ≈ 69 mL (rounded to two significant digits)
Therefore, the volume of the 1.9 mol/L zinc nitrate solution that contains 25.0 g of zinc nitrate is approximately 69 mL when rounded to two significant digits.
