0.218 g of magnesium when dissolved in dilute HCl evolved 218.2 mL of hydrogen at $17^{\circ} C$ and 754.5 mm collected over water.

Question

$0.218 g$ of magnesium when dissolved in dilute $HCl$ evolved $218 \cdot 2 mL$ of hydrogen at $17^{\circ} C$ and $754.5 mm$ collected over water. Find the equivalent weight of the metal. (Vapour pressure of water at $17^{\circ} C =14.4 mm$ ).

Answer 1

Pressure of dry $H _{2}$ :

$\begin{array}{c}(P-f)=(754 \cdot 5-14 \cdot 4) mm =740 \cdot 1 mm \\V=218 \cdot 2 mL ; T=(273+17) K =290 K \\P_{0}=760 mm , V_{0}=? \quad T_{0}=273 K \\\frac{P_{0} V_{0}}{T_{0}}=\frac{(P-f) V}{T}, \therefore V_{0}=\frac{740 \cdot 1 \times 218 \cdot 2}{290} \times \frac{273}{760} mL =200 \cdot 0 mL\end{array}$

Hence equivalent weight of magnesium

$=\frac{\text { wh. of metal }( mg )}{\text { vol. of } H _{2} \text { at S.T.P. }} \times 11200=\frac{0 \cdot 218}{200} \times 11200=12 \cdot 2 \text {. }$