When a current of 0.75 A is passed through `CuSO_(4)` solution of 25 min, 0.369g of copper is deposited at the cathode. Calculate the atomic mass of copper.

When a current of 0.75 A is passed through `CuSO_(4)` solution of 25 m

1.	When a current of 0.75 A is passed through `CuSO_(4)` solution of 25 min, 0.369g of copper is deposited at the cathode. Calculate the atomic mass of copper.
Answer» Quantity of electricity passed `=(0.75A)(25xx60s)=1125C` `Cu^(2+) to 2e^(-)toCu` Thus, 1 mole of Cu (i.e., gram atomic mass) is deposited by 2F, i.e, `2xx96500C` Now, `1125` C deposit Cu=0.369 g `2xx96500C` will deposit Cu=`(0.369)/(1125)xx2xx96500g=63.3g` hence, atomic mass of copper=63.3

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When a current of 0.75 A is passed through `CuSO_(4)` solution of 25 min, 0.369g of copper is deposited at the cathode. Calculate the atomic mass of copper.

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