
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
Answer
233.4k+ views
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.
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.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reactions (2025-26)

