Question

Task 3. Unknown Ca Sample Gravimetric Analysis Result Unknown Ca Stock Solution Used 10.00 Unknown Ca Working Solution Vol 500.00 Unknown Ca Working Solution Aliquot 50.00 Molar Mass of CaC204H2O 146.113 Molar Mass of Ca 40.078 t value at 95% C.LL. 4.3027 Number of Measurements (n) Degree of Freedom (d.f.) (mL) (mL) (mL) (g/mol) (g/mol) #3 Table 3-1. Unknown Ca gravimetric analysis result Replicate # #1 #2 Empty Crucible Mass 25.8233 28.2176 Crucible+ppt. Mass 25.9851 28.3787 Precipitate Mass (g) 0.1618 0.1611 Ca in aliquot (g) 0.0444 0.0442 [Ca] in unknown (mg/L) 44381 44189 Mean (ppm Ca) 44271 SD (ppm Ca) RSD(ppt) 2.2 C.I. (95% C.LL.) 246 28.5178 28.6791 0.1613 0.0442 44244 99 Table 3-2. Summary of unknown Ca gravimeetric analysis conditions Conditdion Operation 45 old precipitate heating time (min) microwave used (old/new) microW heating with 1 or 2 set* post precipitation waiting (d)** ppt. mass confirmed (Y/N) ***

Need help on how to solve to get the Ca in aliquot

Answer 1

This question is based over calculation of mole concept.

solutionis given below:

Since PPT consist CaC2O4H2O .in question we have to find mass of Ca in aliquot.

Suppose suppose mass of calcium is x

Mass of CaC2O4.H2O	mass of Ca
146.113(molecular mass)	40.078(molecular mass)
0.1618(mass of ppt)	X

Now,

X=(0.1618x40.078)/146.113

X=0.044(final answer)

T there is is best possible answer of your question,if you have any doubt then comment please.

Thanks