JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.
Q. Given that the abundances of isotopes $^{54} \mathrm{Fe},^{56}$ Fe and $^{57}$ Fe are $5 \%, 90 \%$ and $5 \%$, respectively, the atomic mass of Fe is :
(A) 55.85 (B) 55.95 (C) 55.75 (D) 56.05
[JEE 2009]
Ans. (B)
$\mathrm{M}_{\mathrm{AVg}}=\frac{\mathrm{M}_{1} \mathrm{X}_{1}+\mathrm{M}_{2} \times \mathrm{X}_{2}+\mathrm{M}_{3} \times \mathrm{X}_{3}}{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}$
$=\frac{5 \times 54+56 \times 90+5 \times 57}{5+90+5}$
= 55.95
Q. Silver (atomic weight $=108 \mathrm{g} \mathrm{mol}^{-1}$ ) has a density of 10.5 g $\mathrm{cm}^{-3}$. The number of silver atoms on a surface of area $10^{-12} \mathrm{m}^{2}$ can be expressed in scientific notation as y ´ $10^{\mathrm{x}}$. The value of x is -
[JEE 2010]
Ans. 7
$\mathrm{d}=\frac{\text { mass }}{\mathrm{V}} \Rightarrow 10.5 \mathrm{g} / \mathrm{cm}^{3}$ means in $1 \mathrm{cm}^{3}=10.5 \mathrm{g}$ of $\mathrm{Ag}$
No of atom of $\mathrm{Ag}$ in $1 \mathrm{cm}^{3}=\frac{10.5}{108} \times \mathrm{N}_{\mathrm{A}}$
in $1 \mathrm{cm}, \mathrm{no}$ of atom of $\mathrm{Ag}=\left(\frac{10.5}{108} \times \mathrm{N}_{\mathrm{A}}\right)^{1 / 3}$
in $10^{-12} \mathrm{m}^{2}$ or $10^{-8} \mathrm{cm}^{2},$ No of atom of $\mathrm{Ag}$
$=\left(\frac{10.5 \mathrm{N}_{\mathrm{A}}}{108}\right)^{2 / 3} \times 10^{-8}=\left(\frac{1.05 \times 6.022 \times 10^{23}}{108}\right)^{2 / 3} \times 10^{-8}$
$=1.5 \times 10^{7}$
Hence $\mathrm{x}=7$
Q. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is
(A) 1.78 M (B) 2.00 M (C) 2.05 M (D) 2.22 M
[JEE 2011]
Ans. (C)
Vol of Solution $=\frac{\text { mass }}{\text { density }}$
$=\frac{1000+120}{1.15} \mathrm{mL}$
$=\frac{1120}{1.15} \mathrm{ml}$
Molarity $=\frac{120 / 60}{\frac{1120}{1.15}} \times 1000=2.05 \mathrm{M}$
Q. A compound $\mathrm{H}_{2} \mathrm{X}$ with molar weight of 80 g is dissolved in a solvent having density of
0.4 g /ml, Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is.
[JEE 2014]
Ans. 8
$\mathrm{V}_{\text {solvent }}=\mathrm{V}_{\text {solution }}$
3.2 mol present in 1L solution (solvent)
3.2 mol / solvent
$1000 \times 0.4=400 \mathrm{gm}$
$=.4 \mathrm{Kg}$
Molality $=\frac{3.2}{0.4}=8 \mathrm{m}$
Q. The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g $\mathrm{cm}^{-3}$. The ratio of the molecular weights of the solute and solvent, $\left(\frac{\mathrm{MW}_{\text {solute }}}{\mathrm{MW}_{\text {solikent }}}\right)$, is
[JEE - ADV. 2016]
Ans. 9
1 mole solution has 0.1 mole solute and 0.9 mole solvent
Let $\mathrm{M}_{1}=$ Molar mass solute
$\mathrm{M}_{2}=$ Molar mass solvent