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Q1.26:
If the density of some lake water is 1.25g \(mL^{-1}\) and contains 92 g of \(Na^{+}\) ions per kg of water, calculate the molarity of \(Na^{+}\) ions in the lake.

Solution :

Step 1: Identifying Given Variables

  • Density of lake water ($\rho$) = $1.25 \, \text{g mL}^{-1}$
  • Mass of solute ($Na^{+}$ ions) ($w_2$) = $92 \, \text{g}$
  • Molar mass of $Na^{+}$ ions ($M_2$) = $23 \, \text{g mol}^{-1}$
  • Mass of solvent ($H_2O$) = $1 \, \text{kg} = 1000 \, \text{g}$

Step 2: Calculating Total Mass and Volume of the Solution

Total mass of the solution ($W_{total}$) = Mass of solute + Mass of solvent

$W_{total} = 92 \, \text{g} + 1000 \, \text{g} = 1092 \, \text{g}$

Using the relationship $Volume = \frac{Mass}{Density}$ [Definition of Density]:

$V_{solution} = \frac{1092 \, \text{g}}{1.25 \, \text{g mL}^{-1}} = 873.6 \, \text{mL}$

To convert to liters: $V = \frac{873.6}{1000} = 0.8736 \, \text{L}$

Step 3: Calculating Number of Moles of Solute

Number of moles ($n_2$) = $\frac{\text{Given mass}}{\text{Molar mass}}$

$n_{Na^+} = \frac{92 \, \text{g}}{23 \, \text{g mol}^{-1}} = 4.0 \, \text{mol}$

Step 4: Calculating Molarity

Molarity ($M$) is defined as the number of moles of solute dissolved per liter of solution [Definition of Molarity]:

$M = \frac{n_2 (\text{mol})}{V_{solution} (\text{L})}$

$M = \frac{4.0 \, \text{mol}}{0.8736 \, \text{L}} \approx 4.5787 \, \text{mol L}^{-1}$

Visual Representation of Solution Parameters

Mass of Solvent = 1000g Mass of Na+ = 92g Total Volume = 0.8736 L Molarity = 4.58 M
Parameter Value
Total Mass 1092 g
Volume 0.8736 L
Moles of $Na^{+}$ 4.0 mol

Final Solution: The molarity of $Na^{+}$ ions in the lake water is 4.58 M.


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