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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2025 · 2 April · Evening Shift · Chemistry · Solutions

When 1 g each of compounds AB and AB_2 are dissolved in 15 g of...

When 1 g each of compounds AB and AB2\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 ____ ×10−1\times 10^{-1} (Nearest integer)
(Given : Molal boiling point elevation constant is 0.5 K kg mol−1)\left.0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)

Correct answer

0

Step-by-step solution

For AB
<br/>ΔTb=2.7 K<br>2.7=1×0.5×m<br> m=275<br/>\begin{aligned}<br />& \Delta \mathrm{T}_{\mathrm{b}}=2.7 \mathrm{~K} \\<br>& 2.7=1 \times 0.5 \times \mathrm{m} \\<br>& \mathrm{~m}=\frac{27}{5}<br />\end{aligned}
Let molar mass of AB=x\mathrm{AB}=\mathrm{x}.
<br/> So 1/x15×1000=275<br>x=12.34<br/>\begin{aligned}<br />& \text{ So } \frac{1 / x}{15} \times 1000=\frac{27}{5} \\<br>& x=12.34<br />\end{aligned}
For AB2\mathrm{AB}_2
<br/>ΔTb=1.5 K<br>1.5=1×0.5×m<br> m=3<br/>\begin{aligned}<br />& \Delta \mathrm{T}_{\mathrm{b}}=1.5 \mathrm{~K} \\<br>& 1.5=1 \times 0.5 \times \mathrm{m} \\<br>& \mathrm{~m}=3<br />\end{aligned}
Let molar mass of AB2=y\mathrm{AB}_2=\mathrm{y}
So 1/y15×1000=3\frac{1 / \mathrm{y}}{15} \times 1000=3
<br/>y=100045<br>y=22.22<br/>\begin{aligned}<br />& y=\frac{1000}{45} \\<br>& y=22.22<br />\end{aligned}
Now let a and b be atomic masses of A and B respectively, then
<br/>A+b=12.34...(i)<br> A+2 b=22.22...(ii)<br> B=22.22−12.34=9.88<br> Now a=12.34−9.88=2.46<br>=24.6×10−1=25×10−1<br/>\begin{aligned}<br />& \mathrm{A}+\mathrm{b}=12.34 \quad...(i)\\<br>& \mathrm{~A}+2 \mathrm{~b}=22.22 \quad...(ii) \\<br>& \mathrm{~B}=22.22-12.34=9.88 \\<br>& \text{ Now } \mathrm{a}=12.34-9.88=2.46 \\<br>& =24.6 \times 10^{-1}=25 \times 10^{-1}<br />\end{aligned}
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