1.

You are supplied with 500 mL each of 2 N HCl and 5 N HCI. What is the maximum volume of 3 M HCI that you can prepare using only these two solutions?

Answer»

250 mL
500 mL
750 mL
1000 mL

Solution : Maximum volume of 3 M HCL solution can be PREPARED by taking 500 mL of 2 N HCl and some volume of 5 N HCl (let x mL). For HCI, Molarity = Normality
So, `N_(1)V_(1) + N_(2)V_(2) = N_(3)V_(3)`
`=500 xx 2 + x xx 5 = (x+500) xx 3`
(`THEREFORE` Volume of 3 M solution formed = 500 mL + x mL)
`therefore 1000 + 5x = 250 mL`
Then, maximum volume of 3 M solution formed
`=500 + 250 = 750 mL`


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