Question

$1g$ of an alloy of aluminum and magnesium when treated with excess of dil.HCl form magnesium chloride and aluminium chloride and hydrogen collected over mercury at $0^{o}C$ has a volume of $1.20L$ at $0.92$ atmospheric pressure. The composition of alloy is:

• A. $Al=54.86\%$ and $Mg=45.14\%$
• B. $Mg=54.86\%$ and $Al=45.14\%$
• C. $Al=45.86\%$ and $Mg=54.14\%$
• D. $Al=45.68\%$ and $Mg=54.32\%$

Answer 1

The correct option is A $Al=54.86\%$ and $Mg=45.14\%$

Let us first calculate the volume of $H_2$ at N.T.P:
Applying: $\frac{p_1{V}_1}{T_1}$ = $\frac{p_2{V}_2}{T_2}$
Where $p_1=0.92atm$ and $p_2=1atm$
$V_1=1.20L$ $V_2=?$
$T_1=273K$ $T_2=273K$
$V_2=\frac{p_1\times V_1\times T_2}{p_2\times T_1}=\frac{0.92\times 1.20\times 273}{(1\times 273)}$
= $1.104L$

Let the mass of $Al$ in alloy = $xg$
and mass of $Mg$ in alloy = $(1-x)g$
According to the reactions:
$2Al$ + $6HCl\to AlC{l}_3+3H_2$
$2\times 27$ $3\times 22.4L$
= $54g$ = $67.2L$

$Mg$ + $2HCl$ $\to$ $MgC{l}_2$ + $H_2$
$24g$ $22.4L$
⇒ $54g$ of $Al$ give $H_2$ at N.T.P = $67.2L$
$xg$ of $Al$ will give $H_2$ at N.T.P = $\frac{(67.2\times x)}{54}$
Similarly, $24g$ of $Mg$ give $H_2$ at N.T.P = $22.4L$
$(1-x)g$ of $Mg$ will give $H_2$ at N.T.P = $22.4\times \frac{(1-x)}{24}$
Total $H_2$ liberated :
= $\frac{67.2\times x}{54}+\frac{22.4\times (1-x)}{24}$ = $1.104$
Solving for $x$, we get
$x=0.5486g$
⇒Mass of $Al$ in the alloy = $0.5486g$
$\%ofAl=\frac{0.5486}{1}\times 100=54.86\%$
and $\%ofMginalloy=100-54.86=45.14\%$