JEE Main 2025 — Chemistry Physical Chemistry

2025integermedium

When 1 g each of compounds AB and $\mathrm{AB}_2$ are dissolved in 15 g of water separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. The atomic mass of A (in amu) is ____ $\times 10^{-1}$ (Nearest integer)

(Given : Molal boiling point elevation constant is $\left.0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$

Official previous-year question

Held on 2 Apr 2025 · Verified 6 Jul 2026.

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Solution

For AB

$\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{b}}=2.7 \mathrm{~K} \

& 2.7=1 \times 0.5 \times \mathrm{m} \

& \mathrm{~m}=\frac{27}{5}
\end{aligned}$

Let molar mass of $\mathrm{AB}=\mathrm{x}$ .

$\begin{aligned}
& \text { So } \frac{1 / x}{15} \times 1000=\frac{27}{5} \

& x=12.34
\end{aligned}$

For $\mathrm{AB}_2$

$\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{b}}=1.5 \mathrm{~K} \

& 1.5=1 \times 0.5 \times \mathrm{m} \

& \mathrm{~m}=3
\end{aligned}$

Let molar mass of $\mathrm{AB}_2=\mathrm{y}$

So $\frac{1 / \mathrm{y}}{15} \times 1000=3$

$\begin{aligned}
& y=\frac{1000}{45} \

& y=22.22
\end{aligned}$

Now let a and b be atomic masses of A and B respectively, then

$\begin{aligned}
& \mathrm{A}+\mathrm{b}=12.34 \quad...(i)\

& \mathrm{A}+2 \mathrm{b}=22.22 \quad...(ii) \

& \mathrm{~B}=22.22-12.34=9.88 \

& \text { Now } \mathrm{a}=12.34-9.88=2.46 \

& =24.6 \times 10^{-1}=25 \times 10^{-1}
\end{aligned}$

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Solution

For AB

$\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{b}}=2.7 \mathrm{~K} \

& 2.7=1 \times 0.5 \times \mathrm{m} \

& \mathrm{~m}=\frac{27}{5}
\end{aligned}$

Let molar mass of $\mathrm{AB}=\mathrm{x}$ .

$\begin{aligned}
& \text { So } \frac{1 / x}{15} \times 1000=\frac{27}{5} \

& x=12.34
\end{aligned}$

For $\mathrm{AB}_2$

$\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{b}}=1.5 \mathrm{~K} \

& 1.5=1 \times 0.5 \times \mathrm{m} \

& \mathrm{~m}=3
\end{aligned}$

Let molar mass of $\mathrm{AB}_2=\mathrm{y}$

So $\frac{1 / \mathrm{y}}{15} \times 1000=3$

$\begin{aligned}
& y=\frac{1000}{45} \

& y=22.22
\end{aligned}$

Now let a and b be atomic masses of A and B respectively, then

$\begin{aligned}
& \mathrm{A}+\mathrm{b}=12.34 \quad...(i)\

& \mathrm{A}+2 \mathrm{b}=22.22 \quad...(ii) \

& \mathrm{~B}=22.22-12.34=9.88 \

& \text { Now } \mathrm{a}=12.34-9.88=2.46 \

& =24.6 \times 10^{-1}=25 \times 10^{-1}
\end{aligned}$

JEE Main 2025 — Chemistry Physical Chemistry

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Solution

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JEE Main 2025 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

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Keep practicing Physical Chemistry