Given:
Step 1: Calculate the mass of the solution
\[ M_{sol} = V_{sol} \times d_{sol} = 500 \, \text{mL} \times 1.25 \, \text{g/mL} = 625 \, \text{g}. \]
Step 2: Calculate the mass of solute
\[ \text{Mass of solute (CuSO}_4\text{)} = M \times V_{sol} \times \text{Molar mass}. \] \[ = 0.2 \times 0.5 \times 159.5 = 15.95 \, \text{g}. \]
Step 3: Calculate the mass of the solvent
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute}. \] \[ = 625 - 15.95 = 609.05 \, \text{g} = 0.60905 \, \text{kg}. \]
Step 4: Calculate the molality
\[ m = \frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}} = \frac{0.1}{0.60905}. \] \[ m = 0.164 \, \text{mol/kg} = 164 \times 10^{-3} \, \text{mol/kg}. \]
Final Answer
The molality of the solution is:
\[ 0.164 \, \text{mol/kg (or } 164 \times 10^{-3} \, \text{mol/kg)}. \]
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32