Molarity is moles of solute per litre of SOLUTION — so the density is there to turn a MASS of lake water into a VOLUME of lake water, and nothing else.The trap in this question is the phrase "per kg of water". The water here
is the lake water — the same lake water whose density \(\displaystyle 1.25\ \mathrm{g\ mL^{-1}}\) you were just handed. So \(\displaystyle 1\ \mathrm{kg}\) of lake water is \(\displaystyle 1\ \mathrm{kg}\) of
solution, containing \(\displaystyle 92\ \mathrm{g}\) of \(\displaystyle \mathrm{Na^{+}}\) dissolved in it.
Step $\displaystyle 1$ — moles of \(\displaystyle \mathrm{Na^{+}}\).The defining relation for amount of substance is
\[n=\frac{m}{M}
\]
where \(\displaystyle n\) is the amount in mol, \(\displaystyle m\) the mass in g, and \(\displaystyle M\) the molar mass in \(\displaystyle \mathrm{g\ mol^{-1}}\). For the sodium ion, \(\displaystyle M(\mathrm{Na^{+}})=23\ \mathrm{g\ mol^{-1}}\) (an electron weighs about \(\displaystyle 1/42000\) of a sodium atom, so losing one changes the molar mass in the fifth decimal place — ignore it).
\[n(\mathrm{Na^{+}})=\frac{92\ \mathrm{g}}{23\ \mathrm{g\ mol^{-1}}}=4.00\ \mathrm{mol}
\]
Aside — per mole, not per gram. Molarity counts particles, not grams. Every gram figure in this question has to pass through \(\displaystyle M\) before it can enter the molarity expression.
Step $\displaystyle 2$ — volume of that $\displaystyle 1$ kg of lake water.Density is mass per unit volume,
\[\rho=\frac{m}{V}\qquad\Longrightarrow\qquad V=\frac{m}{\rho}
\]
with \(\displaystyle \rho\) in \(\displaystyle \mathrm{g\ mL^{-1}}\), \(\displaystyle m\) in g, \(\displaystyle V\) in mL. Substituting the \(\displaystyle 1\ \mathrm{kg}=1000\ \mathrm{g}\) of lake water:
\[V=\frac{1000\ \mathrm{g}}{1.25\ \mathrm{g\ mL^{-1}}}=800\ \mathrm{mL}=0.800\ \mathrm{L}
\]
Aside — the density belongs to the solution. Students often divide by \(\displaystyle 1.00\ \mathrm{g\ mL^{-1}}\) here, reasoning "it's water". It is not pure water; it is a salty solution that is $\displaystyle 25$% denser, and \(\displaystyle 1\ \mathrm{kg}\) of it occupies only \(\displaystyle 800\ \mathrm{mL}\), not \(\displaystyle 1000\ \mathrm{mL}\). Using the wrong density would hand you \(\displaystyle 4\ \mathrm{M}\).
Step $\displaystyle 3$ — molarity.\[M=\frac{n(\text{solute})}{V(\text{solution in L})}
\]
where \(\displaystyle n\) is in mol and \(\displaystyle V\) is the volume of the
whole solution, in litres.
\[M(\mathrm{Na^{+}})=\frac{4.00\ \mathrm{mol}}{0.800\ \mathrm{L}}=5.00\ \mathrm{mol\ L^{-1}}
\]
Rounded once, at the end, to three significant figures (the data — \(\displaystyle 92\ \mathrm{g}\), \(\displaystyle 1.25\ \mathrm{g\ mL^{-1}}\), \(\displaystyle 1\ \mathrm{kg}\) — supports three): \(\displaystyle 5.00\ \mathrm{M}\).
Two things worth knowing about this question.First, "m" versus "M". The printed answer is often typeset as "$\displaystyle 5$ m", which reads as $\displaystyle 5$
molal. It is not: molality is \(\displaystyle \mathrm{mol\ kg^{-1}}\) of
solvent and never uses a density, whereas this calculation used the density to get a volume. The quantity asked for and computed is molarity, \(\displaystyle 5\ \mathrm{mol\ L^{-1}}\), symbol
M. Lower-case m is an OCR/typography slip, not a different result.
Second, the reading that gives \(\displaystyle 4.58\ \mathrm{M}\). If you insist that "kg of water" means \(\displaystyle 1\ \mathrm{kg}\) of pure solvent, then the solution mass becomes \(\displaystyle 1000+92=1092\ \mathrm{g}\), its volume \(\displaystyle 1092/1.25=873.6\ \mathrm{mL}\), and the molarity \(\displaystyle 4.00/0.8736=4.58\ \mathrm{M}\). That reading collapses on its own terms: a solution cannot hold \(\displaystyle \mathrm{Na^{+}}\) without counter-anions, whose mass is not given, so the solution mass would be unknowable and the problem unsolvable. The intended — and the only workable — reading is that the \(\displaystyle 1\ \mathrm{kg}\) is \(\displaystyle 1\ \mathrm{kg}\) of lake water. The clean \(\displaystyle 92/23=4\) exactly, and \(\displaystyle 1000/1.25=800\) exactly, confirm it.
Answer: \(\displaystyle 5.00\ \mathrm{M}\) — the molarity of \(\displaystyle \mathrm{Na^{+}}\) in the lake water is \(\displaystyle 5.00\ \mathrm{mol\ L^{-1}}\). 
Question: NCERT Class XII Chemistry, Chapter $\displaystyle 1$ (Solutions), Q1.21 — freezing-point depression problem to find atomic masses of A and B from molar masses of AB2 and AB4.Method:
1. ΔTf = Kf·m, with m = (w2/M2)/W1(kg). Rearranged: M2 = Kf·w2/(ΔTf·W1).
2. For AB2: w2=1g, ΔTf=2.3K, \(\displaystyle \mathrm{W_{1}}\)=0.020kg → M(AB2) = $\displaystyle 5.1$×$\displaystyle 1$/($\displaystyle 2.3$×$\displaystyle 0.020$) = $\displaystyle 5.1$/$\displaystyle 0.046$ = $\displaystyle 110.87$ g/mol ≈ $\displaystyle 110.9$ g/mol.
3. For AB4: w2=1.0g, ΔTf=1.3K, \(\displaystyle \mathrm{W_{1}}\)=0.020kg → M(AB4) = $\displaystyle 5.1$×$\displaystyle 1$/($\displaystyle 1.3$×$\displaystyle 0.020$) = $\displaystyle 5.1$/$\displaystyle 0.026$ = $\displaystyle 196.15$ g/mol ≈ $\displaystyle 196.2$ g/mol.
4. Set a = atomic mass A, b = atomic mass B: a+2b = $\displaystyle 110.9$ (i); a+4b = $\displaystyle 196.2$ (ii).
5. (ii)-(i): 2b = $\displaystyle 85.3$ → b = $\displaystyle 42.6$ u (more precisely $\displaystyle 42.64$ u using unrounded molar masses).
6. Substitute into (i): a = $\displaystyle 110.9$ - $\displaystyle 85.3$ = $\displaystyle 25.6$ u (more precisely $\displaystyle 25.59$ u).Full worked solution (formatted for KaTeX, with bold lead idea, named formulas, unit tracking, common-mistake asides, ending in Answer: ...) was written out in the response body.
The type of intermolecular force between two substances is decided by what each one IS — non-polar, polar, or ionic — not by memorising a list of names.Three categories cover every pair below: