What is the molar concentration of the b⁻ ions in 0.25 m xb₃(aq)?

XB3(aq) is an ionic compound which will have the releasing of 3 B⁻ ions ions in water for every molecule of XB3 that dissolves.
XB3(s) --> X+(aq) + 3 B⁻(aq)
            [B⁻] = 0.25 mol XB3/1L × 3 mol B⁻ / 1 mol XB3 = 0.75 M
The answer to this question is [B⁻] = 0.75 M

the answer is true.

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