What volume of a 0.540 mol L^-1 NaOH(aq)solution contains 11.5 g of NaOH? A)0.155 L B)0.532 L C)1.88 L D)6.44 L

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First, calculate the moles of NaOH in 11.5 g. The molar mass of NaOH is approximately 40 g/mol, so $ \frac{11.5 g}{40 g/mol} = 0.2875 $ mol. Given the concentration of the solution is 0.540 mol/L, the volume $ V $ containing 0.2875 mol can be found using $ V = \frac{n}{C} = \frac{0.2875 mol}{0.540 mol/L} = 0.532 L $.

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