calibration (1 Viewer)
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[M] = ppm/MM/1000
= 42 / 134.45 / 1000
= 0.00031238378
around 3x10^-4 --> A
EDIT: Actually I believe, looking back at my working the answer is C. The molarity of CuCl2 would be 3.12*10^-4, but because there are 2 Cl ions for every Cu ion the molarity would be twice that hence around 6.24*10^-4, thus C.
What paper is this btw
= 42 / 134.45 / 1000
= 0.00031238378
around 3x10^-4 --> A
EDIT: Actually I believe, looking back at my working the answer is C. The molarity of CuCl2 would be 3.12*10^-4, but because there are 2 Cl ions for every Cu ion the molarity would be twice that hence around 6.24*10^-4, thus C.
What paper is this btw
Last edited:
Roughly 42 ppm Cu2+ is a concentration of 42 mg / L divided by 63.55 g / mol to give
[Cu2+] = 6.6089... x 10-4 mol / L
Since copper(II) chloride is CuCl2, there are two chloride anions for every copper(II) cation, and so
[Cl-] = 2 x 6.6089... x 10-4 = 1.32... x 10-3 mol / L
And the answer is (D).
[Cu2+] = 6.6089... x 10-4 mol / L
Since copper(II) chloride is CuCl2, there are two chloride anions for every copper(II) cation, and so
[Cl-] = 2 x 6.6089... x 10-4 = 1.32... x 10-3 mol / L
And the answer is (D).
wizzkids
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There are many ways to perform this calculation; here is an alternative to @Luukas.2
Determine the mass factions of Cu and Cl in the compound CuCl2.
This compound is 63.5 amu of Cu and 70.9 amu of Cl, so 47.2% Cu and 52.8% Cl by weight.
Now let's look at Solution 2. Reading off the graph, and using the line of best fit (not the data point) it seems to be 43 ppm Cu, therefore it must contain 48 ppm Cl. Hence in 1000 mL of Solution 2 there must be 0.048 g of Cl. To convert this to moles/L we divide by the molecular mass of Cl, and we get 0.00135 m/L.
Determine the mass factions of Cu and Cl in the compound CuCl2.
This compound is 63.5 amu of Cu and 70.9 amu of Cl, so 47.2% Cu and 52.8% Cl by weight.
Now let's look at Solution 2. Reading off the graph, and using the line of best fit (not the data point) it seems to be 43 ppm Cu, therefore it must contain 48 ppm Cl. Hence in 1000 mL of Solution 2 there must be 0.048 g of Cl. To convert this to moles/L we divide by the molecular mass of Cl, and we get 0.00135 m/L.