For this we weigh $\pu{1.015 g}$ of linseed oil which is dissolved in alcohol. 3 drops of phenolphthalein (pH indicator) are added and a solution of potassium hydroxide of concentration $C = \pu{9.5e-3 mol L-1}$ is poured. The indicator turns to $V = \pu{7.4 mL}$

I calculate like that: $$n_\mathrm{acid} = n_\mathrm{OH} = C_\ce{OH}\cdot V_\ce{OH} = 9.5\cdot 10^{-3} \cdot 7.4\cdot 10^{-3} = \pu{1.444e-4 mol}$$

$$\pu{1.444e-4 mol} → \pu{1.015 g}$$ $$\pu{1.423e-4 mol} ← \pu{1 g}$$ $$\implies I_a = m_\ce{KOH} = 1.423\cdot 10^{-4} \cdot 56 = \pu{7.967 mg}$$

But the teacher responds with $I_a = 3.8$. How to solve?

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Unfortunately, the proportion you are using is not correct. Acid value $\mathrm{AN}$ determination procedure is not much different from an ordinary titration with indicator with the only difference that the mass of titrant is normalized to the mass of oil $m_\mathrm{oil}$:

$$\mathrm{AN} = \frac{T\cdot ΔV_\mathrm{eq}}{m_\mathrm{oil}}$$

where $ΔV_\mathrm{eq}$ is equivalent volume (ideally you have to use a difference with an external standard ("spiking solution"), but you probably didn't do that); $T$ is a so-called titer, or basically mass concentration of the titrant:

$$T = CM$$

so that

$$ \begin{align} \mathrm{AN} &= \frac{C(\ce{KOH})\cdot M(\ce{KOH})\cdot ΔV_\mathrm{eq}}{m_\mathrm{oil}}\\ &= \frac{\pu{9.5e-3 mol L-1}\cdot\pu{56.106 g mol-1}\cdot\pu{7.4e-3 L}}{\pu{1.015 g}}\\ &= \pu{3.89 mg g-1} \end{align} $$

Also, a couple of side notes:

  1. Always carry units across all steps and never omit them.
  2. I recommend not to do intermediate calculations if you don't have to. Instead, first find an algebraic equation and only then plug in all the numbers.
  3. Mind significant figures. If the mass of oil is given with 4 significant figures, it's a good idea to use a molecular mass of the titrant with at least 4 sigfigs too.

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