1.0 L of 0.225 M NaCl corresponds to 1.00×0.225 moles.
Let V be the volume of lead(II) chloride.
The number of moles of PbCl2=V(required)×0.750.
2 moles of NaCl will react with 1 mole of lead nitrate.
Hence, 0.225 moles of NaCl will react with 0.1125 moles of lead nitrate.
V×0.750=0.1125
Thus, the volume is 0.15 L or 150 mL.