You are given 5 solutions: Solution A is made by dissolving 3.0g of CuSO 4 , in

ID: 1075261 • Letter: Y

Question

You are given 5 solutions:

Solution A is made by dissolving 3.0g of CuSO4, in 100 mL of water.

Solution B is made by dissolving 6.0g of CuSO4, in 200mL of water.

Solution C is made by dissolving 5.0 g of CuO in 100 mL of H2SO4.

Solution D is made by dissolving 1.0 g of Cu wire in 250mL of HNO3

Solution E is made by diluting 5.0 mL of 0.1 M CuNO3 solution to 50 mL of water.

Assume that theres no change of volume during solution preparation. Put solutions in the order of high to low concentration of Cu ion.

We know that concentration

M=(mass / molarmass)*(1000/ volume in L)

From law of dilution,MV=M'V'

Where

M= molarity of stock solution

V= volume of stock

M' = molarity of dilute solution

V' = Volume of dilute solution

Solution A : M= (3g/(159.6(g/mol))*(1000/100 mL)=0.188 M

Solution B: M=(6g/(159.6(g/mol))*(1000/200 mL)=0.188 M

Solution C: M= (5.0g/(79.5(g/mol))*(1000/100 mL)=0.629 M

Solution D: M=(1g/(63.5(g/mol))*(1000/250 mL)=0.063 M

Solution E: M=M'=(MV)/V'=(0.1M*5.0 mL )/50 mL= 0.01 M

So the order of sasolutionS is : C > B= A> D > E

M=(

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.