Calculate the molarity (mol L−1) of a solution containing 25 g of C2H5OH in 675 mL solution.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.805
Molarity (M) is the number of moles of solute present in one litre of solution:
\( \displaystyle M=\frac{\text{moles of solute}}{\text{volume of solution in L}}=\frac{w_B/M_B}{V} \)
The volume is that of the whole solution, not of the solvent, and it must be converted to litres.
For ethanol, \( \ce{C2H5OH} \):
\( \displaystyle M_B=2(12)+6(1)+16=46\ \mathrm{g\,mol^{-1}} \)
\( \displaystyle n=\frac{25}{46}=0.5435\ \mathrm{mol} \)
\( \displaystyle V=675\ \mathrm{mL}=0.675\ \mathrm{L} \)
\( \displaystyle M=\frac{0.5435}{0.675}=0.805\ \mathrm{mol\,L^{-1}} \)
So the molarity of the solution is 0.805 mol L−1.
Shortcut: With the volume in mL, use one line: \( \displaystyle M=\frac{w_B\times1000}{M_B\times V_{\mathrm{mL}}}=\frac{25\times1000}{46\times675}=\frac{25000}{31050}\approx0.805 \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Expressing Concentration of Solutions – …
