Question

An alkali chloride(X) forms compound(Y) when it is heated with alcoholic KOH. Acidified permanganate oxidation of (Y) forms two moles of carboxylic acid(Z). Heating of the sodium salt of (Z) with soda lime results in the formation of CH4 What is the molar mass of (X) In g mol- unit?​

Answer 1

Answer:

$\\\tt \dfrac{92.5g}{mol}$

Explanation:

Here's the steps as per your question:

X $\longrightarrow$ Y + KOH $\\\TT --{\Delta}\longrightarrow$ 2(CH3-COOH) $\\\tt -NAOH+ CaO \longrightarrow$ CH4

KOH and heat remove the X (halide group) and returns corresponding alkene, then Y must've been an alkene.

Looking at the number of carboxylic acid produced we the alkene must've been,

H3C-CH=CH-CH3  and for this alkene the corresponding haloalkene must've been H3C-CH(Cl)-CH2-CH3.

NaOH addition will return CH3-COONa and addition of CaO+NaOH and heat will do decarboxylation retuning an alkene wich is CH4 here.

We have:

X=H3C-CH(X)-CH2-CH3

Y=   H3C-CH=CH-CH3

Z= CH3-COOH

X can be any halogen here, therefore mass of X will be

$\\\tt \dfrac{(4(12) + 35.5 + 9)g}{mol} = \dfrac{92.5g}{mol}$

Answer 2

Molar mass of alkali chloride(X)

Explanation:

• According to question, Alkali halide X is treated with alcoholic KOH to form Y. The reaction involved is dehydrohalogenation whose major product is less substituted alkene. So, compound Y is a alkene.
• Next compound Y undergo acidified permanganate oxidation and forms 2 moles of carboxylic acid Z. We cam calculate the number of carbon atoms present in alkene by the number of carbon atoms present in 2 moles of carboxylic acid. So, it is clear that  alkene has 4 number of carbons i.e. compound Y is But-1-ene (CH₃CH₂CH=CH₂).
• Sodium salt of Z i.e. CH₃COONa heated with CaO + NaOH and forms CH₄.
• So, Compound X is CH₃-CH₂-CH(Cl)-CH₃.
• Its molar mass of X can be calculated as

        ⇒ (4×12 + 9×1 +35.5)gmol⁻¹

         ⇒92.5gmol⁻¹