Question

0.7 g of Na2​CO3​.xH2​O is dissolved in 100 ml,20 ml of which required 19.8 ml of 0.1 N HCl. The value of x is

Answer 1

Answer:

The balanced chemical equation for the neutralisation reaction of Na2CO3.xH2O with HCl is given by :

Na2CO3.xH2O + 2​HCl →→ 2 NaCl + CO2 + (x+ 1) H2O

Thus  

M1V1n1(HCl)=M2V2n2(Na2CO3.xH2O) where M is molarityV is volume usedn is stoichiometric coefficient in balanced chemical reactionn1 =2 and n2 = 1M1V1n1(HCl)=M2V2n2(Na2CO3.xH2O) where M is molarityV is volume usedn is stoichiometric coefficient in balanced chemical reactionn1 =2 and n2 = 1

Substituting all the values :

0.1×19.82=M2×201Thus M (Na2CO3.xH2O ) = 0.0495 M0.1×19.82=M2×201Thus M (Na2CO3.xH2O ) = 0.0495 M

Thus for ​Na2CO3.xH2O

molarity = 0.0495 M

molar mass (M)= 23××2+12+16××3 + 18 x  = (106 + 18 x) g/mol

Mass used (m) = 0.7 g

Volume of solution = 100 mL = 0.1 L

Molarity = mM×1Volume of solution in L0.0495= 0.7106+18x×10.1106+18x= 0.70.0495×10.1106+18x = 141.4118 x = 35.41x = 35.4118x = 1.97x ~ 2 Molarity = mM×1Volume of solution in L0.0495= 0.7106+18x×10.1106+18x= 0.70.0495×10.1106+18x = 141.4118 x = 35.41x = 35.4118x = 1.97x ~ 2  

Thus value of x is 2 .Thus compound is Na2CO3.2H2O​

Explanation: