Question

# Crystals of washing soda, Na2CO3.10H2O, lose most of their water content when left in an open...

Crystals of washing soda, Na2CO3.10H2O, lose most of their water content when left in an open container.

Na2CO3.10H2O(s)  → Na2CO3.H2O(s)  +  9 H2O(g)

What will be the decrease in mass for 100.0 g of the crystals from this process?

a. 43.4g

b.56.6 g

c. 63.0 g

d. 81.8 g

Molar mass of Na2CO3 = 106 g/mol

Molar mass of H2O = 18 g/mol

So

Na2CO3.10 H2O = 106 + 180 = 286 g/mol

Na2CO3.H2O = 106 + 18 = 124 g/mol

100 g of Na2CO3.10 H2O = 100 g/286 g/mol => 0.3496 moles

1 mol of Na2CO3.10H2O produces 1 mol of Na2CO3.H2O and 9 moles of H2O

Therefore 0.3496 moles of Na2CO3.10H2O could produce 9*0.3496 moles of H2O ( 3.146 moles)

mass of 3.146 moles H2O = 3.146 mol * 18 g/mol => 56.64 g of H2O.

mass of 1*0.3496 moles Na2CO3.H2O = 0.3496 mole*124 g/mol => 43.4 g of Na2CO3

After loosing H2O ,the resulting mass of Na2CO3.H2O is 43.4 g

So Answer is option A) 43.4 g

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