Answer: 0.0826mol
PV=nRT
n=PV/RT
n=(1atm)(2.1L)/(310K)(0.082057L*atm/mol*K)=0.0826mol
6.11% w/v of Cu2+ implies that 6.11 g of Cu2+ is present in 100 ml of the solution
therefore, 250 ml of the solution would have: 250 ml * 6.11 g/100 ml = 15.275 g
# moles of Cu2+ = 15.275 g/63.546 g mole-1 = 0.2404 moles
1 mole of CuCl2 contain 1 mole of Cu2+ ion
Hence, 0.2404 moles of Cu2+ would correspond to 0.2404 moles of CuCl2
Molar mass of CuCl2 = 134.452 g/mole
The mass of CuCl2 required = 0.2404 moles * 134.452 g/mole = 32.32 grams
Answer:
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Answer:
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