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A mixture of Al and Zn weighing 1.67g was completely dissolved in acid and evolved 1.69litre of $H_2$ at NTP.What was the weight of Al in original mixture?

$(a)\;1.25g\qquad(b)\;2.5g\qquad(c)\;1g\qquad(d)\;2g$

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Let 'a'g and 'b'g are weights of Al and Zn in mixture
$a+b=1.67$--------(1)
Meq of Al+Meq of Zn=Meq of $H_2$
$\large\frac{a}{\Large\frac{27}{3}}$$\times 1000+\large\frac{b}{\Large\frac{65}{2}}$$\times 1000=\large\frac{1.69}{\Large\frac{22.4}{2}}$$\times 1000$-----(2)
From equation (1) & (2)
$a=1.25g\qquad Al$
$b=0.42g\qquad Zn$
Hence (a) is the correct answer.
answered Jan 8, 2014 by sreemathi.v
 

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