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trapecia [35]
4 years ago
14

What is the melting point of the substance?

Chemistry
1 answer:
Katena32 [7]4 years ago
3 0
Depends on what the substance is, but I would guess A. Anything above 32 degrees should melt but again it depends on the substance.
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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 $\large \boxed{\textbf{392 u}}$}

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4 years ago
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Answer:

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Explanation:

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Calculate the pH of a buffer prepared by mixing 20.0 mL of 0.10 M acetic acid and 55.0 mL of 0.10 M sodium acetate
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Calculate the pH of a buffer prepared by mixing 30.0 mL of 0.10 M acetic acid and 40.0 mL of 0.10 M sodium acetate.

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