Question

A sample of CaCo3 conatins 3.01 into 10 to the power 23 ions of Ca2+ and (Co3)2-.The mass of the sample is

Answer 1

Here is the solution

Answer 2

Answer:

$we \: know \: that  CaCO_{3}  contains \: 1 \: mole \: of \: both  \: {Ca}^{2 + }   \: and  \: { CO\frac {2 - }{3} }^{} \:  ions \: that \: is \: , 6.02× {10}^{23} \:  ions \: of \: both \: the \: ions.$

$Now, \: since \: the \: number \: of \: ions \: in \: the \: question \: is \: given \: to \: be  \: 3.01× {10}^{23} \:  which \: is \: half \: of  \: 6.02× {10}^{23} , \: hence, \: the \: number \: of \: moles \: will \: also \: get \: halved.$

$Since \: we \: know \: that  \:moles= \frac{Given mass}{Molecular mass} \\ \\ Therefore,  \frac{1}{2} = \frac{Required mass of  CaCO_{3} }{100}$

$Required mass of  CaCO_{3} =50g$

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