Question

What volume of a 1.1 M solution of NaNO₃ contains exactly 0.398 g of the compound?

Answer 1

We have,

Molarity $= 1.1M$

Mass of compound $= 0.398g$

Molar mass of $NaNO_3 = 85gmol^{-1}$

Molarity $= \dfrac{\dfrac{0.398}{85}}{V}$

Hence, V will be calculated.