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2.380 gm of a metal on treatment with nitric acid and subsequent ignition gave 3.022 gm of the metallic oxide. Specific heat of metal is 0.055. Calculate the exact atomic weight. ​

1 Answer

3 votes
To solve this problem, we need to use the law of conservation of mass to relate the mass of the metal to the mass of its oxide, and then use the specific heat of the metal to calculate its atomic weight.

Step 1: Calculate the mass of oxygen in the oxide
3.022 g (mass of oxide) - 2.380 g (mass of metal) = 0.642 g (mass of oxygen)

Step 2: Calculate the moles of metal and oxygen
moles of metal = 2.380 g / atomic weight of metal
moles of oxygen = 0.642 g / molecular weight of oxygen

Step 3: Write the equation for the reaction between the metal and oxygen
M + O2 → MOx

Step 4: Use the law of conservation of mass to set up an equation relating the moles of metal and oxygen
moles of metal = moles of oxygen / x (where x is the number of moles of oxygen per mole of metal oxide)

Step 5: Solve for x
moles of metal = moles of oxygen / x
2.380 g / atomic weight of metal = 0.642 g / (16 + atomic weight of metal)
2.380 (16 + atomic weight of metal) = 0.642 atomic weight of metal
38.08 + 2.380 atomic weight of metal = 0.642 atomic weight of metal
1.738 atomic weight of metal = 38.08
atomic weight of metal = 21.93

Therefore, the exact atomic weight of the metal is 21.93.
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