If 80 g of copper sulphate CuSO_{4} cdot 5H_ | vedclass

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(C) $1$. Calculate the molar mass of $CuSO_{4} \cdot 5H_{2}O$: $M = 63.54 + 32 + (4 	imes 16) + 5 	imes (2 	imes 1 + 16) = 63.54 + 32 + 64 + 90 = 2

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$64$

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(C) $1$. Calculate the molar mass of $CuSO_{4} \cdot 5H_{2}O$: $M = 63.54 + 32 + (4 	imes 16) + 5 	imes (2 	imes 1 + 16) = 63.54 + 32 + 64 + 90 = 249.54 \ g \ mol^{-1}$.$2$. Calculate the number of moles of $CuSO_{4} \cdot 5H_{2}O$:$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{80 \ g}{249.54 \ g \ mol^{-1}} \approx 0.3206 \ mol$.$3$. Calculate the molarity of the solution:$Molarity = \frac{n}{V(L)} = \frac{0.3206 \ mol}{5 \ L} = 0.06412 \ mol \ L^{-1}$.$4$. Express in the form $x 	imes 10^{-3} \ mol \ L^{-1}$:$0.06412 = 64.12 	imes 10^{-3}$.Thus,the value of $x$ is $64$.

If $80 \ g$ of copper sulphate $CuSO_{4} \cdot 5H_{2}O$ is dissolved in deionised water to make $5 \ L$ of solution. The concentration of the copper sulphate solution is $x \times 10^{-3} \ mol \ L^{-1}$. The value of $x$ is .... .
[Atomic masses: $Cu = 63.54 \ u, S = 32 \ u, O = 16 \ u, H = 1 \ u$]

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If $80 \ g$ of copper sulphate $CuSO_{4} \cdot 5H_{2}O$ is dissolved in deionised water to make $5 \ L$ of solution. The concentration of the copper sulphate solution is $x \times 10^{-3} \ mol \ L^{-1}$. The value of $x$ is .... .[Atomic masses: $Cu = 63.54 \ u, S = 32 \ u, O = 16 \ u, H = 1 \ u$]