Question

A number of moles of water in $488\mathrm{gram}$ of ${\mathrm{BaCl}}_2·2{\mathrm{H}}_2\mathrm{O}$ is-

Answer 1

Step 1: Given data

${\mathrm{W}}_{({\mathrm{BaCl}}_2·2{\mathrm{H}}_2\mathrm{O})}=488\mathrm{g}$ (W is the given mass)

${\mathrm{M}}_{({\mathrm{BaCl}}_2·2{\mathrm{H}}_2\mathrm{O})}=244\mathrm{g}$ (M is the molar mass of the compound)

Step 2: Introducing the formula for calculating the number of moles:

$\mathrm{n}=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}}$

Step 3: Putting the values in the above-mentioned equation to obtain the number of moles of hydrated Barium Chloride

$\mathrm{n}=\frac{488}{244}\mathrm{mol}=2\mathrm{mole}$

Step 4: Calculating the number of moles of water in 2 moles of Barium Chloride

1 mole of Barium Chloride has 2 moles of water associated with it.

Therefore, with 2 moles of Barium Chloride 4 moles of water will be associated with it.

Therefore, the number of moles of water in $488\mathrm{gram}$ of ${\mathrm{BaCl}}_2·2{\mathrm{H}}_2\mathrm{O}$ is 4.