The number of moles of water in 488 grams of ${\mathrm{BaCl}}_2.2{\mathrm{H}}_2\mathrm{O} \mathrm{are}$

Molar mass of ${\mathrm{BaCl}}_2.2{\mathrm{H}}_2\mathrm{O} \mathrm{is} 244 \mathrm{g}.$

244g of ${\mathrm{BaCl}}_2.2{\mathrm{H}}_2\mathrm{O}$ contains 2 moles of  ${\mathrm{H}}_2\mathrm{O}$

488 g ${\mathrm{BaCl}}_2.2{\mathrm{H}}_2\mathrm{O}?$

No. of moles of ${\mathrm{H}}_2\mathrm{O}$ present in given sample =$\frac{2 \mathrm{mol}}{244\mathrm{g}}\times 488g = 4moles$

So, there are 4 moles of ${\mathrm{H}}_2\mathrm{O}$ present in 488 grams of ${\mathrm{BaCl}}_2.2{\mathrm{H}}_2\mathrm{O}$