total number of sulphate ions present in 3.92g of chronic sulphate
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In Chromium Sulphate, Cr2(SO4)3, there are 2 moles of chromium ions and 3 moles of sulphate ions
We know, No. of moles = MassMolar Mass
Molar Mass of Cr2(SO4)3 = 392 g/mol
Therefore, no. of moles in 3.92 g of Cr2(SO4)3 = 3.92 g392 g/mol = 0.01 moL
So, no. of moles of Chromium ions = (2 x 0.01) moles = 0.02 mol
No. of moles of Sulphate ions = (3 x 0.01) moles = 0.03 mol
Total no. of ions/molecules of sulphate = no. of moles x Avagadro number=0.03 x 6.023 x 1023 = 0.18069 x 1023 = 1.8069 x 1022
Hope this answer helps u
We know, No. of moles = MassMolar Mass
Molar Mass of Cr2(SO4)3 = 392 g/mol
Therefore, no. of moles in 3.92 g of Cr2(SO4)3 = 3.92 g392 g/mol = 0.01 moL
So, no. of moles of Chromium ions = (2 x 0.01) moles = 0.02 mol
No. of moles of Sulphate ions = (3 x 0.01) moles = 0.03 mol
Total no. of ions/molecules of sulphate = no. of moles x Avagadro number=0.03 x 6.023 x 1023 = 0.18069 x 1023 = 1.8069 x 1022
Hope this answer helps u
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this is the correct answer and is certified
In Chromium Sulphate, Cr2(SO4)3, there are 2 moles of chromium ions and 3 moles of Sulphate ions
We know, No. of moles =Molar Mass
Molar Mass of Cr2(SO4)3 = 392 g/mol
Therefore, no. of moles in 3.92 g of Cr2(SO4)3 = 3.92 g392 g/mol = 0.01 mol
So, no. of moles of Chromium ions = (2 x 0.01) moles = 0.02 mol
No. of moles of Sulphate ions = (3 x 0.01) moles = 0.03 mol
Total no. of ions/molecules of Sulphate = no. of moles x Avogadro number=0.03 x 6.023 x 1023 = 0.18069 x 1023 = 1.8069 x 1022
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