Answer:
1.5L of NaNO3 must be present
Explanation:
That contains 200g of sodium nitrate. Round to 2 significant digits
To solve this question we need to convert the mass of NaNO3 to moles using its molar mass (85g/mol). With the moles and the molar concentration we can find the volume in liters of the solution:
Moles NaNO3:
200g * (1mol / 85g) = 2.353 moles NaNO3
Volume:
2.353 moles NaNO3 * (1L / 1.60moles) =
1.5L of NaNO3 must be present
