Question

1 x 10²² molecules of CuSO₄.5H₂O was added to water and 100 g solution was obtained. % by weight of CuSO₄ .5H₂O in the solution is?

Answer 1

Answer:

no. of molecules of $CuSO_{4} .5H_{2} O$= $10^{22}$

no. of moles of $CuSO_{4} .5H_{2} O$=$\frac{10^{22} }{6.023X10^{23}}$

Given mass of $CuSO_{4} .5H_{2} O$=$\frac{10^{22} }{6.023X10^{23}}X (molecular mass)= \frac{10^{22} }{6.023X10^{23}}X 160$

                                             =$26.56 X 10^{-1} = 2.656$

%by weight= $\frac{2.656}{100} X100 = 2.656%$%