When I did it, I got like 7%, so i gues the answer is E, is it true.
(i am not sure on this one, but it should be E)
(i am not sure on this one, but it should be E)
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Maybe I'm missing something, but here's how I would go about calculating this.
70.9 mL * 1L/1000 mL * 0.201 mol AgNO3/L * 1 mol Ag+/ 1 mol AgNO3 * 107.8 g / 1 mol Ag+ * 1/5.00 G * 100 = 30.7 %
Answer: E, none of the above.
If all of the chloride in a 5 gram sample of an unknown metal chloride is precipitated as AgCl with 70.9mL of 0.201 M AgNO3, what is the percentage of chloride in the sample?
A. 50.55%
B. 20.22%
C. 1.43%
D. 10.10%
E. none of the above
70.9 mL * 1L/1000 mL * 0.201 mol AgNO3/L * 1 mol Ag+/ 1 mol AgNO3 * 107.8 g / 1 mol Ag+ * 1/5.00 G * 100 = 30.7 %
Answer: E, none of the above.
If all of the chloride in a 5 gram sample of an unknown metal chloride is precipitated as AgCl with 70.9mL of 0.201 M AgNO3, what is the percentage of chloride in the sample?
A. 50.55%
B. 20.22%
C. 1.43%
D. 10.10%
E. none of the above
B
BodybldgDoc
I cannot confirm what the correct answer would be but the answer to the problem is is E none of the above. From the looks of it, i think REH found the percentage of Ag instead of Cl in the sample. So,
1-.307=.693 or 69.3% of Cl in the sample after precipitation. What do ya think?
1-.307=.693 or 69.3% of Cl in the sample after precipitation. What do ya think?
