1.

What volume of 4M HCI and 2M HCI should be mixed to get 500 mL of 2.5M HCI?

Answer»

Let the volume of 4M HCl required to prepare 500 mL of 2.5 M HCI = x mL

Therefore, the required volume of 2M HCI = (500 – x) mL

We know from the equation x = \(\frac{250}{2}\) = 125 mL 

Hence, volume of 4M HCI required = 125 mL

 Volume of 2M HCl required = (500 – 125) mL = 375 mL



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