Question

1. A student following the procedure of this experiment obtained the following data for solutions of Cr(III) ion at 580 nm: Standard Solution Number Volume, in mL 0.1000 M Cr(III) Ion Solution Volume, in ml Distilled Water Molarity of Standard Solution Absorbance of Standard Solution blank 0 10.0 0.000 1.0 9.0 0.0921 0.180 8.0 7.0 6.0 0.278 4.0 0.375 5.0 5.0 0.460 (a) Calculate the molarity of each of the standard solutions.

Answer 1

Question :- Answer :- we know that with increasing volume So, the relation between volm Solution de rases mcreasing vinme concentration of between volume and concentration relation (in molarity Vd solation TM= v Molarity of volume on , Mi = 12M₂ Standard solention invember. Blank : OmL of 0.100 M ( m ion Solution + 10 mL Distilled water so, Molarity of standard 501" Standard Solution cum im solntion is Le canse no 0.00M is • Number -1:- 1 ml of 0.100M ( m ion. Somtion + 9 mL Distilled water W = 1 ml M = 0.100 M ME "2 = total volume My = Molanity of = (1 ml + 9m2) = 10 mL standard sointion . Di M V , M 1 moX X 0.100M 10 nok = 0.010 M 50, Molarity of Standard solution 0.010 m. • Number - 2! mL of 0.100M cr(AI) on solntion V, -amL M = 0.100 M 2 & 8 mL Distilled water = (2+8) mL = 10 ml, 2 V, M. = V₂ M2 on. M₂ = 1 2m x 0.1001 10 m 0.020 M = Molarity of Standard Solution is 0.020 M

• Number - 3 : 3 ml 0.100 M or (nl) ion Solution V = 3 m2, Mr = 0.100 M + 7 mL Dishled water 2 = (3 + time - 10 mL, M₂ = ? VM = 1₂M2 3 mL X 0.200M = 0.030 M 10 AL Molarity of standard solution is 0.030M Number : - 4 mL 0.100M cm (m) lon Solntion t 6 mL Distilled water V = 4 mL, My = 0.100M = 14+6) mL = 10 ml , M2 = ? w My = V₂ M₂ - 4 AL x 0.100M - 10 not 0.040 M standard solution is 0.040M Molarity of numbers - 5:- 5 mL 0.100M Cro (1) ian solation + 5 mL Distilled water w = 5 mL, M, = 8.100 M M = 2 2 = (5 +5/ml = 10 mL NM, = V2 M2 or. M₂= h = SW x 0 10010.050 M 20m Molarity of standard solution is 0.050 M.
We Know that with increasing volume , concentration decreases and decreasing volume , concentration increases.so they are inversely proportional.Thus we can calculate the molarity of standard solution.