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When 0.3 g of a metal is dissolved in di...

When 0.3 g of a metal is dissolved in dilute HCl the volume of `H_(2)` gas liberated is 110 mL at `17^(@)C` and 755 mm Hg of pressure. [Aqueous tension at `17^(@)C=14.4` mm Hg] Determine the equivalent mass of the metal.

Text Solution

Verified by Experts

Pressure of dry hydrogen gas=755-14.4
=740-6 mm Hg.
If the volume of `H_(2)` gas produced at STP be V mL, then
`(110xx740.6)/((273+17))=(Vxx760)/(273)`
So, `V=(110xx740.6xx273)/(290xx760)mL=100.91mL`
Now, mass of 22400 mL of `H_(2)` at STP=`2xx1.008g`
`therefore` Mass of 100.91 mL of `H_(2)` at STP=`(2xx1.088xx100.91)/(22400)g`
=0.00908g
So, 0.00908g of `H_(2)` is displaced by 0.3g of metal.
`therefore` 1.008g of `H_(2)` is displaced by `(0.3xx1.008)/(0.00908)=33.3g` of metal
`therefore` Equivalent mass of the metal=33.3
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