Answer
$138torr$
Work Step by Step
We can calculate the required vapor pressure of hexane as follows:
$m_{C_{10}H_8}=12.35g$
$m_{C_6H_{14}}=100g-12.35g=87.65g$
The number of moles of naphthalene $n_{C_{10}H_8}=\frac{12.35g \space C_{10}H_8\times 1mol\space C_{10}H_8}{128.1702g\space C_{10}H_8}=0.096356mol\space C_{10}H_8$
and the number of moles of hexane $n_{C_6H_{14}}=\frac{87.65g\space C_6H_{14}\times 1mol C_6H_{14}}{86.1748g\space C_6H_{14}}=1.0171molC_6H_{14}$
Now $P=X_{C_6H_{14}}P^{\circ}_{C_6H_{14}}$
$\implies P=(\frac{n_{C_6H_{14}}}{n_{C_6H_{14}}+n_{C_{10}H_8}})P^{\circ}_{C_6H_{14}}$
We plug in the known values to obtain:
$P=(\frac{1.0171mol}{1.0171mol+0.096365mol})(151torr)=138torr$
