Compute the number of moles of aluminium sulphate which contains 10²³SO₄^–2 ions.

Solution

because 3 SO subscript 4 to the power of negative 2 end exponent space are space present space in space one space formula space unit space of space aluminium space sulphate comma space Al subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3 because 10 to the power of 23 SO subscript 4 superscript 2 minus end superscript space ions space would space be space present space in space 10 to the power of 23 over 3 formula space units space of space Al subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3. Now comma space space space space space 6.02 space cross times space 10 to the power of 23 space formula space units space of space Al subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space identical to space 1 space mole space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space

therefore space 10 to the power of 23 over 3 space formula space units space of space Al subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3

identical to space space fraction numerator 1 over denominator 6.02 space cross times space 10 to the power of 23 end fraction cross times 10 to the power of 23 over 3 equals space 5.54 space cross times space 10 to the power of negative 2 end exponent space mole

Some More Questions From Some Basic Concepts of Chemistry Chapter

For Daily Free Study Material Join wiredfaculty