Question

40. A student was trying to measure 20 ml of a solution using the pipette. However, he did not notice that he has introduced air bubbles while filling the pipette. His colleague double-checked the volume and he found out that the volume dispensed was actually 18.5 ml instead of 20 ml. What is the percentage error of such a measurement? (2 points) Bonus Questions 41. H.PO, reacts with LiOH in a neutralization reaction. A standard solution of LiOH is prepared and its molarity is 0.7 M. If 33 ml of LiOH are required to neutralize 100 ml of H3PO4, determine the molarity of H PO (2 points) H2PO4 + LiOH → LiPO4 + H20 (unbalanced). 42. A hydrated salt that is made of Lithium, Chlorine and Oxygen was heated until all water was removed. It was found that the water content of the hydrated salt is 33.62005%. An elemental analysis found that the composition of the other elements in the hydrated salt was: 4.36300 % Lithium 22.12665 % Chlorine 39.8903 % Oxygen (3 points) a. Calculate the empirical formula of this hydrated salt: Li.Cl:0;.?H2O. b. How do you ensure that all the water was evaporated from a hydrated salt?

Answer 1

- 20 ml = 18.5 ml Solution - 9. N0.40;- Volume to be dispensed Volume actually dispensed Now, percentage error is given | (Volume to be dispensed - Volume actually dispucellence to be dispensed % Error Volume 20 -18.5 X100 20 % Error - 1.5 x100 20 y Error : 7.50% 09.No.413- The given equation is LigPO4 + H₂O The balanced equation is given H3PO4 + Lion H₂PO4 + 3 lion Li_PO4 + 3H2O Molarity of Lion = 0.7 m Volume - 33ml = 0.033L. 4 0.7 X 0.033 = con required 0.0231 moles a's moles of Lions It is clear from the above equation, that 3 moles of Lou react with H₂Pou I mole.

I moles. I mole of Lion reacts with H₂PO4 = => 0.0231 moles of Lion react with H2POU Ix00231 = 0.0077 moles 3 í Moles of MyP04 = 0.0077 moles. volume of H₂PO4 used = 100 ml = 0.12 Moles of H3PO4 → Molsity of My PO4 = Volume of H₂PO4 0.0077 0.077 M 코 D.No.42: Celven percentages are. % H₂O 33.62005 % Li 4.36300 % Cl 22-12665 %. 0 and to 89.8903 % 30, we have 4 species, i.ey H20, L1, CI and O. Let us assume that we have 100 og of compound, then masses of all these species will be H₂O 33.62005g = 4.86300g. = 22.126657 - 39.8903 g. CI 0

الله Molar mars 18 glmol. 33. 62005 2 1.867 Mazo 18 Molar mars 4:36300 6194 0.628 Now, let is calculate the No.of moles of all the species present. 4 И,0 ... Moles of H₂O = of li = 6.94 -94 g/mol. Moles of Lis 5.5 g/mol. Moles of cez of o .: Moles of De = no Dividing all the low of moles by the least No of moles, sie, by moles of al Molar mars cm El = 35.5 22.12665 35.5 0.623 nal Molor mars 16 g/mol- 39.8903 2.493 16 MH2O ทน na no n ci nei nei nel 0:628 1867 0.623 ) 0.623 0.623 ) 0.623 2.493 0-623 1 20947 4.001 → ） ) Rounding off these values to the nearest intergeht. Integer - 3) 4 - 120 Li CI 0 So, correct empoieal formula is LI, CI, 043420 of LiC104:3420