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0.3g of chloroplatinate of an organic diacidic base left 0.09g of platinum on ignition. The molecular weight (in g/mol) of the organic base is:
A. 120
B. 240
C.180
D. 60

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Last updated date: 18th Jun 2024
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Answer
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Hint: We have the given formula: $\dfrac{weight\text{ }of\text{ }chloroplatinate}{weight\text{ }of\text{ }platinum}=\dfrac{equivalent\text{ }weight\text{ salt}}{equivalent\text{ }weight\text{ }platinum}$using this formula and substituting given values of different quantities in it we get the equivalent weight of the given compound. Also, equivalent weight is known as the mass of one equivalent, i.e. the mass of a substance taken under consideration which will combine with or displace a definite quantity of another substance in a chemical reaction.

Complete step by step answer:
- In this question we have to find the molecular weight of the organic base.
- First note down the given information in the question:
Weight of chloroplatinate = 0.3 g
Weight of platinum left = 0.09 g
- And we know should know the atomic mass
Atomic mass of Platinum = 195 g
Atomic mass of chlorine = 35.5 g
- Atomic mass is required to find equivalent mass.
-In order to find the molecular mass of the base we will be equating the given weight of chloroplatinate and platinum to equivalent weight of salt and platinum.
$\dfrac{weight\text{ }of\text{ }chloroplatinate}{weight\text{ }of\text{ }platinum}=\dfrac{equivalent\text{ }weight\text{ salt}}{equivalent\text{ }weight\text{ }platinum}$
-Now to solve this let us assume that the original base is B.
-So according to the question, the chemical equation will be
     $2B+{{H}_{2}}PtC{{l}_{6}}\to {{B}_{2}}{{H}_{2}}PtC{{l}_{2}}\underrightarrow{\Delta }Pt$
-The equivalent weight of ${{B}_{2}}{{H}_{2}}PtC{{l}_{2}}$ will be = 2B+ 2(Atomic mass of platinum) +6(mass of chlorine)
                         = 2B + $2\times 195$+ $6\times 35.5$= 2B +410
-Now we know that
 $\dfrac{weight\text{ }of\text{ }chloroplatinate}{weight\text{ }of\text{ }platinum}=\dfrac{equivalent\text{ }weight\text{ salt}}{equivalent\text{ }weight\text{ }platinum}$
As we discussed above
Therefore,
$\dfrac{0.3}{0.09}=\dfrac{2B+410}{195}$
After rearranging and solving
 $2B=\dfrac{0.3}{0.09}\times 195-410$
Therefore, equivalent weight of B = 120
We can find the molecular weight of the base by the following formula:
\[molecular\text{ }weight\text{ }=\text{ }equivalent\text{ }weight~\times ~acidity\]
= $120\times 2$= 240.
Hence the molecular weight of the organic base is option B. 240 g/mol.

Note:
-\[Equivalent\text{ }weight\text{ }=\dfrac{Molecular\text{ }mass}{valency}\]
-Before putting values in the formula check that all the quantities are in SI unit.