a certain amount of a metal whose equivalent mass is 28 displaces 0.7L of H2 at STP from an acid hence mass of the element is.
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Answered by
13
----I hope it helpfull for u soo plzzz marks as brillant answer. equvalent weight of metal will be equivalentnweight of H2
no of mole pf h2 formed =pv/rt =1×.7/273×.0821 =.0312
let the weight of metal be x
x/28=.0312/2
x=1.747 g
Anonymous:
PLZZ MARKS AS BRILLANT ANSWER.
thnx
Why reported
sorry i am new i didnt know
but u solved this ans in a diff way thnx
Welcome
we can not solve this question by ideal gas law.because this is case of acid mixtures
I hope it helfull for u soo plzzz marks as brillant answer
Answered by
28
0.7 L of H2 at STP = 0.7 /22.4 = 1/32 moles = 1/16 grams
mass of 'x' displaces ---- 1/16 gram of H2 ??
Equiv mass of 28 displaces ... 1 gram of H2
x = 28/16 = 1.75 gms
mass of 'x' displaces ---- 1/16 gram of H2 ??
Equiv mass of 28 displaces ... 1 gram of H2
x = 28/16 = 1.75 gms
At STP conditions one mole (molecular mass in grams) of an ideal gas occupies 22.4 liters.... so at STP 2 grams of Hydrogen occupies 22.4 liters...
Then 0.7 L of H2 will weigh 1/6 grams..
i hope it is now easier to understand the calculations
this is nice but i have another good method also..thanx
Yougot your answer that's good
what is that method ? please explain that here. i want to know, if there is another method... write that here ... at least the formula or name of the method
on equating gram equivalents of metal and H2gas.ie.m/28=0.7×2/22.4,m=28×0.7×2/22.4,then on solving we would get m=1.75,
Yeah
ya it is correct ans.but u tried well sujitrck
Ohh thnx
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