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