4.) An excess of a divalent metal M was dissolved in a limited volume of hydrochloric
acid. If 576 cm of hydrogen were liberated
at s. t.p, what was the mass of metal that
produced this volume of hydrogen. (M= 24,
H= 1, molar volume of a gas at s. t.p = 22.4
dm}
beib 10​

To make the problem more interesting, let's assume that you don't know the formula for dilution calculations. The idea with diluting a solution is that the number of moles of solute will remain constant after the initial solution is diluted. The only thing that changes in such cases is the volume of the solution. This means that if you know how many moles of solute you have in the target solution, you also know how many moles of solute were present in the stock solution sample. Use the molarity and volume of the target solution to determine how many moles of hydrochloric acid, HCl, you need in that solution

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