Question
Option 3 : 65.2%
Detailed Solution
सही उत्तर 65.2% है।
prepsnap.in • Part -AQuestion:
A $0.500$ g sample of impure NaCl gives $0.800$ g of AgCl precipitate on treatment with excess AgNO3. The percentage purity of NaCl in the sample is: (AgCl = $143.32$, NaCl = $58.44$) / अशुद्ध NaCl के $0.500$ g नमूने से आधिक्य AgNO3 द्वारा $0.800$ g AgCl अवक्षेप मिलता है। नमूने में NaCl की प्रतिशत शुद्धता है: (AgCl = $143.32$, NaCl = $58.44$)
Correct Answer: 65.2%
Moles of AgCl = $0.800/143.32 = 0.005581$ mol, equal to moles of NaCl precipitated. Mass of NaCl = $0.005581 \times 58.44 = 0.3262$ g, so purity = $0.3262/0.500 \times 100 = 65.2$%. / AgCl के मोल = $0.800/143.32 = 0.005581$ मोल, जो अवक्षेपित NaCl के मोलों के बराबर है। NaCl का द्रव्यमान = $0.005581 \times 58.44 = 0.3262$ g, अतः शुद्धता = $0.3262/0.500 \times 100 = 65.2$%।