calculate number of if ion present in 5.85 g of sodium chloride
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Answer:
1.2042×10^23
Explanation:
5.85 gm of NaCl=5.85/58.5=0.1 moles
Each NaCl particle is equivalent to one Na^+ one Cl^-
2 ions
total moles of ions =0.1×2=0.2 moles
no of ions=0.2×6.022×10^23=1.2042×10^23.
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=
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