Chemistry - Mole Concept Question with Solution | TestHub

Given weight of A₃B₂C₅ compound = $x$ g.

Weight of A-atoms = $Y$ g.

Weight of B-atoms = $Z$ g.

Gram-atoms (moles) of C-atoms = $U$.

So, we can find the weight of C-atoms:

Weight of C-atoms = Weight of A₃B₂C₅ – (Weight of A + Weight of B)

$=x-Y-Z$

Since moles of C-atoms = $\frac{WeightofC-atoms}{MolarmassofC-atom}$,

$$U=\frac{x-Y-Z}{MolarmassofC-atom}$$

Therefore, Molar mass of C-atom = $\frac{x-Y-Z}{U}$.

So, R → (4).

Now, let us consider the molar mass of the compound = $M$.

For A₃B₂C₅,

Moles of C-atoms = $U$.

$$\frac{x}{M}\times 5=U$$

$$M=\frac{5x}{U}=Molarmassofcompound$$

So, S → (3).

Now, moles of B = $\frac{Z}{M_B}$.

$$\frac{x}{(5x/U)}\times 2=\frac{Z}{M_B}$$

$$\frac{2U}{5}=\frac{Z}{M_B}$$

$$M_B=\frac{5Z}{2U}=AtomicweightofB$$

So, Q → (5).

Now, moles of A = $\frac{WeightofA}{MolarmassofA}=\frac{Y}{M_A}$.

$$\frac{x}{(5x/U)}\times 3=\frac{Y}{M_A}$$

$$\frac{3U}{5}=\frac{Y}{M_A}$$

$$M_A=\frac{5Y}{3U}$$

So, P → (1).

Answer:

P → 1

Q → 5

R → 4

S → 3