Find the mass of Ag around the Nickel pole

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SUMMARY

The discussion focuses on calculating the mass of silver (Ag) deposited around a nickel (Ni) pole when immersed in a 50 ml AgNO3 solution with an initial concentration of 0.1 mol/liter, reducing to 0.02 mol/liter. The number of moles of Ag is determined to be 0.004 moles, derived from the difference in initial and final concentrations. The mass of Ag is calculated using the formula m = n * M, where M is the molar mass of silver (108 g/mol), resulting in a mass of 0.432 grams. The final answer confirms the correct approach but emphasizes the need for clarity between grams and moles.

PREREQUISITES
  • Understanding of molarity and concentration calculations
  • Knowledge of stoichiometry and chemical equations
  • Familiarity with the concept of moles in chemistry
  • Basic skills in unit conversion between grams and moles
NEXT STEPS
  • Study stoichiometric calculations in chemical reactions
  • Learn about molarity and its applications in solution chemistry
  • Explore the properties and reactions of silver nitrate (AgNO3)
  • Investigate the electrochemical processes involving nickel and silver
USEFUL FOR

Chemistry students, educators, and anyone involved in chemical calculations or electrochemistry experiments will benefit from this discussion.

prishila
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Homework Statement


A nickel pole is put in an AgNO3 solution with volume 50ml and concentration 0.1 mol/litre until the concrentration is 0.02 mole/litre. Find the mass of Ag that is put around the Nickel pole.

Homework Equations


2AgNo3+Ni->Ni(NO3)2+2Ag
number of moles= concrentration*volume
n1=0.1*0.05=0.005 moles
n2=0.02*0.05=0.001 moles.
n1-n2=0.004 moles.
So the numebr of Ag moles will be 0.004 too

The Attempt at a Solution


n1=0.1*0.05=0.005 moles
n2=0.02*0.05=0.001 moles.
n1-n2=0.004 moles.
So the number of Ag moles will be 0.004 too
Then m=0.004*108=0.432 moles. Am I right?
 
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prishila said:
m=0.004*108=0.432 moles

Grams, not moles. Other than that what you did looks OK.