calculate number of cations present in 58.5g of sodium chloride.Atomic mass of sodium is 23,chlorine is 35.5
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Cation is ion with +ve charge. Here it is Na.
The atomic mass of NaCl is 58.5 u (Na = 23 u + Cl = 35.5 u). Therefore, we have 1 mole of NaCl.
In NaCl 1mole Na reacts with 1 mole Cl.
Therefore 1 mole Na is present.
By Avagrado's number, 1 mole of any substance contains 6 x 10^23 particles.
Therefore, 6 x 10^23 cations are there.
The atomic mass of NaCl is 58.5 u (Na = 23 u + Cl = 35.5 u). Therefore, we have 1 mole of NaCl.
In NaCl 1mole Na reacts with 1 mole Cl.
Therefore 1 mole Na is present.
By Avagrado's number, 1 mole of any substance contains 6 x 10^23 particles.
Therefore, 6 x 10^23 cations are there.
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