Question

Answer any or all please, I'm using the answers to check my accuracy, thank you :)

TUIL Paragraph Styles 7.45 mL of 0.12 M sodium carbonate (Na2CO3) 8. 250 mL of 0.75 M lithium nitrite (LINO2) 9. 56 mL of 1.1 Miron (II) phosphate (Fe3(PO4)2) 10. 6.7 L of 4.5 M ammonium nitrate (NH4NO3)

Answer 1

To Calculate grams of solute required to prepare required solution

= Molecular weight of compound x molarity x Volume in litre

Molecular weight of Na2CO3 = 105.98

solute required to prepare = 105.98*0.12*0.045= 0.572 gm

Molecular weight of LiNO2 = 52.94

solute required to prepare = 52.94*0.75*0.25 = 9.92 gm

Molecular weight of Fe3(PO4)2 = 357.48

solute required to prepare = 357.48*1.1*0.056 = 22.02 gm

Molecular weight of NH4NO3 = 80.04

solute required to prepare = 80.04*4.5*6.7 = 2413.2 gm