The value of y is \!\! \!\! . .

Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Advanced 2021 — Paper 1 — Question 39

The value of $\mathbf{y}$ is $\text{ }\!\!~\!\!\text{ }$ . .

Numerical answer — enter this value.

Molar mass of `U' = 122 g or 100 g

\text{P (0.1 mole)} \xrightarrow{100\%} \text{S (0.1 mole)}

2S \xrightarrow[\text{(80\%)}]{\text{Aldol condensation}} T \xrightarrow[\left(\tfrac{0.1}{2} \times 0.8\right)]{80\%} U \left(\tfrac{0.1}{2} \times 0.8 \times 0.8\right)

So, mass of `U' =

\frac{0.1}{2} \times 0.8 \times 0.8 \times 100 = 3.20 \, \text{gm}

\text{Mass of `U'} = \frac{0.1}{2} \times 0.8 \times 0.8 \times 122 = 3.90 \, \text{gm}

Answer key and solution verified before publishing.

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