The number of ions present in 2 L of a solution of 1.6 M Al2(SO4)3 is; [Given: NA = 6 × 1023 ]
4.8 × 1022
4.8 × 1023
9.6 × 1024
9.6 × 1022
No. of molecules of Al2(SO4)3 = 1.6 ×2× 6 ×1023
No. of ions = 5 × 1.6 × 2 × 6 × 1023 =9.6 × 1024
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