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