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How many grams of CuCl2 are required to prepare 1250 G of a 1.255% CuCl2 solution,

How many grams of CuCl2 are required to prepare 1250 G of a 1.255% CuCl2 solution,

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Answer #1

mass \, percentage = \frac{mass of solute}{mass of solution}\times100

Given,

mass percentage = 1.255%

mass of solution = 1250gm

subsituting these values in the above equation,

mass \, percentage = \frac{mass of solute}{mass of solution}\times100\\\\1.255 = \frac{mass of solute}{1250}\times100\\\\ mass of solute =15.6875gm

15.6875gm is the amount of CuCL2 needed to prepare a 1.255% of 1250gm of solution.

lets see how thus make sense ,if we have a 1250gm of solution then a 10% of somwthing would amount for 125gm in thaat solution and similarly a 1% will amount for 12.5 gm in that solution . as we have to make a 1.255% our answer came out to be near to 1% that is 15.6875gm

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