Question

# 4.215 g of a metallic carbonate was heated in a hard glass tube and the CO2 evolved was found to measure 1336 ml at 27 and 700 mm pressure. What is the equivalent weight of the metal?

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Solution

## Step 1: Given data :${P}_{1}=760mm\phantom{\rule{0ex}{0ex}}{P}_{2}=700mm$${V}_{1}=?{V}_{2}=1336mL\phantom{\rule{0ex}{0ex}}{T}_{1}=273K,{T}_{2}=27+273=300k$Step 2: Formula used ${\mathrm{P}}_{1}{\mathrm{V}}_{1}/{\mathrm{T}}_{1}={\mathrm{P}}_{2}{\mathrm{V}}_{2}/{\mathrm{T}}_{2}$ ----------------------1)Step 3: Calculation of VolumeFrom 1) $760×{V}_{1}/273=700×1336/300\phantom{\rule{0ex}{0ex}}{V}_{1}=7×1336×273/3×760\phantom{\rule{0ex}{0ex}}=1119.7mL\phantom{\rule{0ex}{0ex}}=1.119L$Step 4: Evaluating Weight of Carbon dioxide Therefore $1.119LC{O}_{2}=4.215g$ metallic carbonate$22.4\mathrm{L}{\mathrm{CO}}_{2}$ is given by $\frac{4.215×22.4}{1.119}g$ metallic carbonate $=84.37g$Step 5: Calculating Equivalent weight of metal Metal carbonate is ${\mathrm{MCO}}_{3}=\mathrm{M}+12+48\phantom{\rule{0ex}{0ex}}=\mathrm{M}+60$Atomic weight of Metal $\mathrm{M}=84.3-60\phantom{\rule{0ex}{0ex}}=24.3$Equivalent weight of metal =$\frac{Molecularweight}{2}$$=\frac{24.3}{7}\phantom{\rule{0ex}{0ex}}=12.18$

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