Q.When 0.1 mol $CoCl_3(NH_3)_5$ is treated with excess of $AgNO_3$, 0.2 mol of $AgCl$ are obtained. The conductivity of solution will correspond to

Concept. In a solution simultaneously saturated with two sparingly soluble salts sharing a common ion, both solubility products must be satisfied by the same common-ion concentration. Here CO$_3^{2-}$ is common to MgCO$_3$ and Ag$_2$CO$_3$.

Step 1 — carbonate concentration from MgCO$_3$.

Step 2 — silver-ion concentration from Ag$_2$CO$_3$.

Step 3 — take the square root.

which is the option written as $\sqrt{1.6}\times10^{-5}$ M — option (D).

ANSWER: D

Both salts share the carbonate ion. From the $\text{MgCO}_3$ equilibrium:

For $\text{Ag}_2\text{CO}_3$, $K_{sp}=[\text{Ag}^+]^2[\text{CO}_3^{2-}]$, so

The key here is that Cu$^{2+}$ oxidises I$^-$ to I$_2$ while itself being reduced to Cu$^+$, which immediately precipitates as insoluble CuI. This is a classic redox-precipitation reaction, and it is the basis of the iodometric estimation of copper:

For every 2 mol of Cu$^{2+}$ that reacts, 4 mol of I$^-$ are consumed — so each mole of Cu$^{2+}$ consumes 2 mol of I$^-$.

1. Find moles of Cu$^{2+}$ added.

   Volume = 100 mL = 0.1 L, concentration = 0.05 M.

   Moles of Cu$^{2+}$ = $0.1 \times 0.05 = 0.005$ mol.

2. Find initial moles of I$^-$ (from KI).

   Volume of KI solution = 1 L, concentration = 0.1 M.

   Moles of I$^-$ initially = $1 \times 0.1 = 0.1$ mol.

3. Determine moles of I$^-$ consumed.

   From the balanced equation, 2 mol of I$^-$ is consumed per mol of Cu$^{2+}$:

   Moles of I$^-$ consumed = $2 \times 0.005 = 0.01$ mol.

4. Find remaining moles of I$^-$.

   Remaining I$^-$ = $0.1 - 0.01 = 0.09$ mol.

5. Find total volume of the resultant solution.

   Volume of Cu$^{2+}$ solution = 0.1 L, volume of KI solution = 1 L.

   Total volume = $0.1 + 1 = 1.1$ L (volumes assumed additive).

6. Calculate final concentration of KI (equal to the concentration of remaining I$^-$, since K$^+$ is a spectator ion).

   Concentration = $\frac{\text{moles of I}^-\text{ remaining}}{\text{total volume}} = \frac{0.09}{1.1}$ mol L$^{-1}$.

The key idea is that for a weak electrolyte like formic acid, the molar conductivity $\Lambda_m$ at a finite concentration is directly proportional to the degree of dissociation $\alpha$, because only the dissociated fraction contributes to carrying current. The limiting molar conductivity $\Lambda_m^0$ (at infinite dilution) represents the conductivity if the acid were fully dissociated. So we simply scale it down by $\alpha$.

Why this works: Kohlrausch’s law tells us that $\Lambda_m^0$ for an electrolyte is the sum of the limiting conductivities of its ions. For a weak acid, the actual molar conductivity at concentration $c$ is $\Lambda_m = \alpha \Lambda_m^0$, because the number of charge carriers is reduced by the factor $\alpha$. This is a direct consequence of the definition of degree of dissociation.

Now, step by step:

1. Find the limiting molar conductivity $\Lambda_m^0$ of formic acid. Using Kohlrausch’s law:

Given $\lambda^0(\mathrm{H}^+) = 349.6~\mathrm{S~cm^2~mol^{-1}}$ and $\lambda^0(\mathrm{HCOO}^-) = 54.6~\mathrm{S~cm^2~mol^{-1}}$,

2. Use the degree of dissociation. The degree of dissociation $\alpha$ is given as $11.0\% = 0.110$. For a weak electrolyte,

So,

3. Check the units. The given ionic conductivities are in $\mathrm{S~cm^2~mol^{-1}}$, so the result is naturally in the same units. That matches option (B). Option (A) gives the same numerical value but in $\mathrm{S~m^2~mol^{-1}}$, which would be $10000$ times smaller (since $1~\mathrm{S~m^2~mol^{-1}} = 10^4~\mathrm{S~cm^2~mol^{-1}}$). So (A) is incorrect. Options (C) and (D) are half the value, which would correspond to $\alpha = 5.5\%$, not $11\%$.

The concept first

Molar conductivity $\Lambda_m = \dfrac{\kappa}{c}$ measures how well one mole of electrolyte conducts. As you dilute a solution, $\Lambda_m$ rises, because the ions get further apart and interfere with one another less. It cannot rise forever, though — it approaches a ceiling, the limiting molar conductivity $\Lambda_m^0$, the value at infinite dilution where each ion moves as if the others did not exist.

The trouble is that you can never actually measure infinite dilution. So Kohlrausch's great empirical discovery matters enormously: for a strong electrolyte (fully dissociated — KCl, HCl, $\mathrm{NaCl}$…), the approach to that ceiling is linear in the square root of concentration:

where $A$ depends on the solvent, temperature and the charge type of the electrolyte. The $\sqrt{c}$ dependence comes from Debye–Hückel theory: each ion is surrounded by an oppositely charged "ionic atmosphere" whose drag scales as $\sqrt{c}$.

That equation lets you get the un-measurable quantity by extrapolation: plot $\Lambda_m$ against $c^{1/2}$, draw the straight line, and read where it crosses the axis.

Step-by-step

1. Map the equation onto $y = mx + c$.

The graph's axes are exactly $y = \Lambda_m$ and $x = [\mathrm{KCl}]^{1/2}$, so this mapping is direct.

2. Interpret the intercept. Setting $x = c^{1/2} = 0$ means $c = 0$ — infinite dilution. Then $\Lambda_m = \Lambda_m^0$. The $y$-intercept IS the limiting molar conductivity.

3. Read it off. The dashed line in the figure meets the $y$-axis precisely at the tick labelled 150. Therefore

4. What is the "$m = 90$" doing there? It is the magnitude of the slope, i.e. the constant $A$ in Kohlrausch's equation. It characterises how fast $\Lambda_m$ falls with concentration; it is not used to compute $\Lambda_m^0$. The options $\frac{150}{90}$ and $150\times 90$ exist only to tempt you into combining the two numbers — there is no equation in which those combinations mean anything. Option (B), 90, mistakes the slope for the intercept.

5. Sanity check against real data. The accepted $\Lambda_m^0$ for KCl in water at $298\ \mathrm{K}$ is about $149.9\ \mathrm{S\,cm^2\,mol^{-1}}$ — the graph is drawn from the real number.

6. Note also the negative slope. The line falls to the right, consistent with $\Lambda_m$ decreasing as concentration increases — exactly what the minus sign in $-A c^{1/2}$ demands, and a quick check that the graph is being read the right way round.

ANSWER: D

The key is to recall the definitions of these two physical quantities, not just memorise units. Surface tension arises from cohesive forces at a liquid’s surface — it’s the force needed to hold the surface together per unit length. Viscosity measures a fluid’s resistance to flow; it relates shear stress to velocity gradient.

1. Surface tension

   • Definition: force per unit length acting along the surface.
   • Force is measured in newtons (N), length in metres (m).
   • So its SI unit is $\mathrm{N\ m^{-1}}$.
   • (Equivalently, since $1\ \mathrm{N} = 1\ \mathrm{kg\ m\ s^{-2}}$, we could also write $\mathrm{kg\ s^{-2}}$, but the problem uses N.)

2. Dynamic viscosity

   • Definition: from Newton’s law of viscosity, $\tau = \eta \frac{du}{dy}$, where $\tau$ is shear stress (force per area) and $\frac{du}{dy}$ is velocity gradient.
   • Shear stress unit: $\mathrm{N\ m^{-2}}$ (or $\mathrm{Pa}$).
   • Velocity gradient unit: $\mathrm{s^{-1}}$ (since velocity in $\mathrm{m/s}$ divided by distance in $\mathrm{m}$).
   • Rearranging: $\eta = \frac{\tau}{du/dy}$ has units $\frac{\mathrm{N\ m^{-2}}}{\mathrm{s^{-1}}} = \mathrm{N\ s\ m^{-2}}$.
   • In base SI units: $1\ \mathrm{N\ s\ m^{-2}} = 1\ \mathrm{kg\ m\ s^{-2} \cdot s \cdot m^{-2}} = \mathrm{kg\ m^{-1}\ s^{-1}}$.

3. Matching to options

   • Surface tension: $\mathrm{N\ m^{-1}}$
   • Viscosity: $\mathrm{kg\ m^{-1}\ s^{-1}}$ (or equivalently $\mathrm{N\ s\ m^{-2}}$)
   • Option (B) lists exactly these: $\mathrm{N\ m^{-1}}$ for surface tension, then $\mathrm{kg\ m^{-1}\ s^{-1}}$ for viscosity.