if 0.01 mole of  Na₂CO₃ is required , amount of Na₂CO_{3 .}10H_​2O to be taken is???

Dear Student,

The equation for dissociation of the sodium bicarbonate decahydrate is:
$$\begin{gathered} N{a}_{2}C{O}_3.10H_{2}O\to N{a}_{2}C{O}_3 + 10H_{2}O \\ x y z \end{gathered}$$, where x,y and z are the weights of the compounds.

Mole ratio of Na₂CO₃:H_​2O = 1:10
Now, number of moles of Na₂CO₃ = 0.01
Therefore, number of moles of H₂O = 0.01 x 10 =0.1

Now, 

y = weight of Na₂CO₃ = Number of moles x Molar Mass = 0.01 x 106 = 1.06 g
z = weight of H₂O = 0.1 x 18 = 1.8 g

Therefore, x = y + z = 1.06 + 1.8 = 1.86 g

Thus, 1.86 g of Na₂CO₃.10H₂O are required.

Regards