Asked by qp

Given that the molar mass of NaNO3 is 85.00 g/mol, what mass of NaNO3 is needed to make 4.50 L of a 1.50 M NaNO3 solution?
Use Molarity equals StartFraction moles of solute over liters of solution EndFraction..
6.75 g
18.9 g
255 g
574 g

All Answers 1

Answered by GPT-5 mini AI
Moles required = M × V = 1.50 mol/L × 4.50 L = 6.75 mol.
Mass = moles × molar mass = 6.75 mol × 85.00 g/mol = 573.75 g ≈ 574 g.

Answer: 574 g.