What volume of toluene (C7H8) must be added to 1.0 mL of ethyl acetate (C4H8O2) to make an equimolar mixture, given densities 0.898 g/mL (ethyl acetate) and 0.867 g/mL (toluene)?

What “equimolar” means here

An equimolar mixture has the same number of moles of each component. So we find the moles in 1.0 mL of ethyl acetate, then add enough toluene to match that mole amount.

Convert 1.0 mL ethyl acetate to moles

First convert volume to mass using density:

$$m_{EA} = (1.0\,\text{mL})(0.898\,\text{g mL}^{-1}) = 0.898\,\text{g}$$

Molar mass of ethyl acetate $C_4H_8O_2$:

$$M_{EA} = 4(12.011) + 8(1.008) + 2(15.999) \approx 88.11\,\text{g mol}^{-1}$$

Now moles:

$$n_{EA} = \frac{0.898}{88.11} \approx 0.01019\,\text{mol}$$

Find the toluene volume that gives the same moles

Set $n_{tol} = n_{EA} = 0.01019\,\text{mol}$.

Molar mass of toluene $C_7H_8$:

$$M_{tol} = 7(12.011) + 8(1.008) \approx 92.14\,\text{g mol}^{-1}$$

Required toluene mass:

$$m_{tol} = n_{tol}M_{tol} = (0.01019)(92.14) \approx 0.939\,\text{g}$$

Convert mass to volume using density $\rho_{tol} = 0.867\,\text{g mL}^{-1}$:

$$V_{tol} = \frac{m_{tol}}{\rho_{tol}} = \frac{0.939}{0.867} \approx 1.08\,\text{mL}$$

So you should add about $1.08\,\text{mL}$ of toluene to 1.0 mL of ethyl acetate.