Using ΔTb=Kb×Mw×W1000
For PQ: 1.176=5×MPQ1×501000=MPQ100
MPQ=1.176100=85 g/mol
For PQ2: 0.689=MPQ2100
MPQ2=0.689100=145 g/mol
P+Q=85 ... (i)
P+2Q=145 ... (ii)
Subtracting: Q=60 g/mol
From (i): P=85−60=25 g/mol
JEE Main 2026 — Chemistry Physical Chemistry
Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ2. When 1 g of PQ is dissolved in 50 g of solvent ′A′,ΔTb was 1.176 K while when 1 g of PQ2 is dissolved in 50 g of solvent ′A′,ΔTb was 0.689 K. (Kb of ' A′=5 K kg mol−1). The molar masses of elements P and Q (in gmol−1) respectively, are :
Held on 21 Jan 2026 · Verified 6 Jul 2026.
70,110
65,145
25,60
60,25
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?
  Which of the following point in Figure 2 most accurately represents the nodal surface as shown in Figure 1?
At 298 K, the mole percentage of $\mathrm{N}_{2}(\mathrm{~g})$ in air is $80 \%$. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of $\mathrm{N}_{2}(\mathrm{~g})$ in water at 298 K ? ($\mathrm{K}_{\mathrm{H}}$ for $\mathrm{N}_{2}$ is $6.5 \times 10^{7} \mathrm{~mm} \mathrm{Hg}$)
Electricity is passed through an acidic solution of $\mathrm{Cu}^{2+}$ till all the $\mathrm{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is $\_\_\_\_$ mL. (Nearest integer) [Given: $\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) \mathrm{E}_{\text {red }}^{\mathrm{o}}=+0.34 \mathrm{~V}$ $\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \mathrm{E}_{\text {red }}^{\mathrm{o}}=+1.23 \mathrm{~V}$ Molar mass of $\mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of $\mathrm{O}_{2}=32 \mathrm{~g} \mathrm{~mol}^{-1}$ Faraday Constant $=96500 \mathrm{C} \mathrm{mol}^{-1}$ Molar volume at $\mathrm{STP}=22.4 \mathrm{~L}$]
Arrange the following resultant mixtures in increasing order of their pH values A. $10$ mL $0.2$ M Ca(OH)$_2$ + $25$ mL $0.1$ M HCl B. $10$ mL $0.01$ M H$_2$SO$_4$ + $10$ mL $0.01$ M Ca(OH)$_2$ C. $10$ mL $0.1$ M H$_2$SO$_4$ + $10$ mL $0.1$ M KOH Choose the correct answer from the options given below:
Work through every JEE Main Physical Chemistry PYQ, year by year.