Calculate the molarity of a solution made by

Chemistry

1 answer:

Kazeer [188]3 years ago

4 0

Answer:

0.1933 M

Explanation:

Data provided in the question:

Number of Moles of Na₂SO₄ initially taken = 0.145 mol

Volume of solution, V = 750 mL = 0.750 L

Now, the molarity is given as:

Molarity = $\frac{\textup{moles of solute}}{\textup{volume of solution in liter}}$

Molarity = $\frac{0.145}{0.750}$

Molarity = 0.1933 M

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Suppose that 98.0g of a non electrolyte is dissolved in 1.00kg of water. The freezing point of this solution is found to be -0.4

Sholpan [36]

Answer:

$\large \boxed{\text{392 u}}$

Explanation:

1. Calculate the molal concentration

The formula for the freezing point depression by a nonelectrolyte is

$\Delta T_{f} = -K_{f}b\\b = -\dfrac{\Delta T_{f}}{ K_{f}} = -\dfrac{-0.465 \, ^{\circ}\text{C}}{\text{1.86 $\, ^{\circ}$C$\cdot$kg$\cdot$mol}^{-1}} = \text{0.250 mol/kg}$

2. Calculate the moles of solute

$\begin{array}{rcl}b & = & \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}\\\\\text{moles of solute} & = & b \times {\text{kilograms of solvent}}\\n & = &\text{0.250 mol/kg} \times \text{1.00 kg}\\ & = & \text{0.250 mol}\\\end{array}$

3. Calculate the molecular mass

$\begin{array}{rcl}\text{Moles} & = &\dfrac{\text{mass}}{\text{molar mass}}\\\\\text{0.250 mol} & = & \dfrac{\text{98.0 g}}{MM}\\\\MM & = & \dfrac{\text{98.0 g }}{\text{0.250 mol}}\\\\& = &\textbf{392 g/mol}\\\end{array}\\\text{The molecular mass of the solute is $\large \boxed{\textbf{392 u}}$}$