How many moles are there in 6.37 x 10(-2) L of 0.0832 M barium nitrate?

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Answer 1 · 2014-07-28T18:20:43

There are 5.30 × 10⁻³ mol of barium nitrate.

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In your problem, volume = 6.37 × 10⁻² L and molarity = 0.0832 mol/L. So,

Moles = 6.37 × 10⁻² L × #(0.0832"mol")/(1"L")# = 5.30 × 10⁻³ mol.