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What volume of CCl4 having density 1.5g/cc contains 1 * 10 to the power 25 chlorine atoms
Dear student,
number of chlorine atoms in ccl4 = 4
number of moles of ccl4 = 10^25/(4*6.623*10^23) = 3.775 moles
weight of 3.775 moles = 3.775*[12+4*35] = 573.8
volume required = 573.8/1.5 =382.33 litres
i hope this clears :)
thanks for the answer. but i did not understand why we multiply 4 with avogadro number
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