Question

How many aluminum ions are contained in 0.0128 uL of a 2.56 * 10 ^-4 M solution of Al(NO3)3

Answer 1

contained solution of How many aluminium ions are in 0.0128 uL of a 2.56 x Lot M AlCNO3)3 Answer = Number of aluminium ion 0.197 X 10 to con solution . for solving This problem, first we calculate mole of ALCNO3)3 using Molarity formula 'After calculating mole of Al(NO3)3 , we can calculate mole of Aluminium ion. Then we multiply mole of Aluminium ion by avogadro's Number. so we get value of aluminium ion. Volume of AICNO3)3 solution = 0.0128 ML = 0.0128x10 °L {iful = 1062 Molarity of solution - 2.56 x 10-4 M since we know. Molarity of solution : mole of the AlcNoz), Volume of solution (in L) but all the value and calculat mole of ALCNO3)3

2.56 X 10 4 M = mole of AICNO3)3 0.0128 x 10-6 L mole of AlCNO3)2 = 2.56* 10 4 x 0.0128 x 16'mole mole of AICNO3)3 = 0:032768x 10-10 mole -() Now we write chemical reaction of ALCNO3)3 in ionic form which tells how the mole of Aluminium ion and Nitrate in related to mole of AlCNO3)₂ Al(NO3)3 = Alt + 3N03 This show that I mole of AICNO3)3 contains 1 mole of Als + Caluminium ion) and 3 mole of NO 3 (nitratto Ion). so we can can say mole of aluminiuton Ion = mole of Al(NO3). mole of aluminium Ion = 0.032768 x 10 10 mole (from equation (i)

since we know I mole contains 6.023 X 10 23 atom which is equal to avogadro'd number so 1 mole of aluminium ion contains 6.023x 10 con 0.032768X 10 - 10 mole of aluminium ion contains = 6.023 X 102X0.032768 nlod.com = 0.197 X 1013 ion 2.3 This means o.1978 1ot ion of aluminium present in 0.032768 X 10 - 10 mole of aluminium ion. so I final answer = 0.197 X 10 aluminium ion. 13
summary:

Answer = 0.197×10^10 ion

Solution:

First we calculate mole of Al(NO3)^3 using molarity formula.

Mole of Al(NO3)3 = molarity of solution × volume of solution ( in L)

Since volume of solution has unit microlitre. So we first changed the unit of volume in litre .

1uL = 10^ -6 L

After calculating mole of Al(NO3)^3 , we can calculate mole of aluminum ion which is equal to mole of Al(NO3)^3.

After calculating mole of Aluminium ion , we multiply mole of Aluminium ion by 6.023 ×10^23. So we get number of Aluminium ion.