Q.

10.0 mL of Na2CO3 solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings:4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL, Based on these readings, and convention of titrimetric estimation the concentration of Na2CO3 solution is ___mM (Round off to the Nearest Integer)

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answer is 50.

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Detailed Solution

Na2CO3+2HCl→    2NaCl+CO2+H2On1=1    n2=1m1=?    m2=0.2v1=10    v2=510×m11=5×0.22     m1=0.05 m1=50×10−3
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10.0 mL of Na2CO3 solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings:4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL, Based on these readings, and convention of titrimetric estimation the concentration of Na2CO3 solution is ___mM (Round off to the Nearest Integer)