Question

The density of \(3 M\) solution of \(NaCl\) is \(10 \,g \,mL -1\). Molality of the solution is _______ \(\times 10^{-2} m\) (Nearest integer). Given: Molar mass of Na and \(Cl\) is \(23\) and \(355 \,g \,mol ^{-1}\) respectively.

Hint:

Molality depends only on the mass of the solvent and is independent of tempera ture, unlike molarity.

Answer 1

Correct Answer: 364

The molality of a solution can be calculated using the formula:

Molality (m) = Moles of solute / Mass of solvent in kg

1. Step 1: Calculate the mass of 1 L of solution
   The density of the solution is given as 1.0 g/mL, so the mass of 1 L of the solution is:
   Mass of solution = 1.0 g/mL × 1000 mL = 1000 g
2. Step 2: Calculate the mass of NaCl in 1 L solution
   The molarity of the solution is 3 M, which means there are 3 moles of NaCl in 1 L of the solution.
   The molar mass of NaCl is:
   Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
   Mass of NaCl = 3 mol × 58.5 g/mol = 175.5 g
3. Step 3: Calculate the mass of the solvent (water)
   The mass of the solvent is the total mass of the solution minus the mass of NaCl:
   Mass of solvent = 1000 g − 175.5 g = 824.5 g = 0.8245 kg
4. Step 4: Calculate the molality
   Molality (m) = 3 mol / 0.8245 kg = 3.64 mol/kg
   Expressing this in the form of ×10⁻²:
   Molality = 364 × 10⁻² m

Answer 2

The correct answer is 364

$m=\frac{1000\times M}{1000\times d-M\times M.W\text{of solute}}$
$=\frac{1000\times 3}{1000\times 1-(3\times 58.5)}=3.64$
$=364\times 10^{-2}$