Question

What weight of $CuS{{O}_{4}}.5{{H}_{2}}O$ must be taken to make $0.5$ litre of $0.01M$ copper $(II)$ ion solution?

Answer 1

Molarity is defined as the number of moles of substance present per litre of the solution. Number of moles of the substance is defined as the ratio of the weight of the substance to its molecular weight. Molecular weight describes the amount which is present in one mole of the substance.

Complete Step By Step Answer:
Molarity of the $CuS{{O}_{4}}.5{{H}_{2}}O$ solution is given as
$Molarity = \dfrac{{Number\,of\,CuS{O_4}.5{H_2}O}}{{Volume\,of\,solution\,(in\,litres)}} - - - - \left( 1 \right)$
Number of moles of $CuS{{O}_{4}}.5{{H}_{2}}O$ is given as
$Number\,of\,moles = \dfrac{{Weight\,of\,CuS{O_4}.5{H_2}O}}{{Molecular\,weight\,of\,CuS{O_4}.5{H_2}O}} - - - - \left( 2 \right)$
The given values are $molarity=0.01M,\text{Volume of Solution}=0.5litre$
Put these value in equation (1), then we get
$\text{No.of moles}=0.005$
Molecular weight of $CuS{{O}_{4}}.5{{H}_{2}}O$ is given as
$Mol.weight\,ofCuS{{O}_{4}}.5{{H}_{2}}O=At.weightofCu+At.weightofS+9\times (At.weightofO)+10\times (At.weightofH)$ $Mol.weight\,ofCuS{{O}_{4}}.5{{H}_{2}}O=249.5g$
Put the above values in equation (2), then we get
$weight\,ofCuS{{O}_{4}}.5{{H}_{2}}O=1.2475g$
Hence, $1.2475g$ of $CuS{{O}_{4}}.5{{H}_{2}}O$ must be taken to make $0.5$ litre of $0.01M$ copper $(II)$ ion solution.

Note:
It is important to note that $1.2475g$ of $CuS{{O}_{4}}.5{{H}_{2}}O$ must be taken to make $0.5$ litre of $0.01M$ copper $(II)$ ion solution. The molecular weight of $CuS{{O}_{4}}.5{{H}_{2}}O$ comes out to be $249.5g$ . The number of moles of $CuS{{O}_{4}}.5{{H}_{2}}O$ is $0.005$ .