the number of sulphate ion present in 3.92 g of chromium sulphate​

Dear Student

In Chromium Sulphate, Cr​2(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 ​Cr​2(SO4)3 = 392 g/mol
Therefore, no. of moles in 3.92 g of ​Cr​2(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

Regards
 

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