Kindly provide the answer of this qn - Chemistry -

Question

Kindly provide the answer of this qn

Hrishi Kumar Sharma · Accepted Answer

Dear students, The water of crystallization will be 10H2O 500 ml of
2M HCl +100ml of 2M H2SO4Â  Total milliequivalents of
Hydronium ionÂ willÂ be 500x2+100x4 {since
H2SO4 will give 2 H+} 1400milliequivalents = 1 4 equivalents The
equivalentsÂ of OH- added in form of Monobasic alkali=1
Net after neutralization will be 0 4 equivalentsÂ of
Hydronium ion 30ml of this solution required 20 ml of 143 gm Na2CO3
xH2O inÂ one-litreÂ solution Now just equate
both milliequivalents We know, 2H+ + Na2CO3===>Â 
2NaOH + CO2 THAT MEANS 2 HYDRONIUM IONS ARE NEEDED FOR EACH Na2CO3
MOLECULE n factor for Na2CO3 is 2 Molarity of Na2CO3 for one liter
=143g/(106+18x) Normality=Mxn factor =2M 0 4x30mL=Â 
Â 2x143g/(106+18x) x20mL x=20 6 Regards