what mass in grams of a molecular substance (mole weight=50.0g/mol) must be added to 500g of water to produce a solution that boils at 101.56(degrees)C?

To determine the mass of the molecular substance required, we need to use the formula:

\( \Delta T_b = k_b \cdot m \cdot I \)

where:
\( \Delta T_b \) = boiling point elevation
\( k_b \) = molal boiling point elevation constant
\( m \) = molality of the solution
\( I \) = van't Hoff factor (for a molecular substance, I = 1)

First, we need to calculate the molality (\( m \)) using the following formula:

\( m = \frac{n_{\text{solute}}}{m_{\text{solvent}}} \)

Given:
Molecular weight of the substance = 50.0 g/mol
Mass of water = 500 g

We can calculate the number of moles of the solute using the following formula:

\( n_{\text{solute}} = \frac{m_{\text{solute}}}{M_{\text{solute}}} \)

Substituting the given values into the formula:

\( n_{\text{solute}} = \frac{m_{\text{solute}}}{50.0} \)

Since we are assuming that the solute completely dissolves in water, the number of moles of the solute is equal to the number of moles of particles present. Hence, the van't Hoff factor (\( I \)) is 1 for molecular substances.

Now, let's calculate the boiling point elevation (\( \Delta T_b \)). The boiling point elevation is the difference between the boiling point of the solution and the boiling point of the pure solvent. The boiling point of water, \( T_b\), is 100.00°C.

\( \Delta T_b = T_b(\text{solution}) - T_b(\text{solvent}) = 101.56 - 100.00 = 1.56 \text{°C} \)

Finally, we can use the formula to calculate the mass of the solute required:

\( \Delta T_b = k_b \cdot m \cdot I \)
\( 1.56 = k_b \cdot \frac{n_{\text{solute}}}{m_{\text{water}}} \)

Since the molal boiling point elevation constant for water is 0.512 °C/m, we can substitute the values and solve for \( m_{\text{solute}} \):

\( 1.56 = 0.512 \cdot \frac{m_{\text{solute}}}{500} \)

Now we can solve for \( m_{\text{solute}} \):

\( m_{\text{solute}} = \frac{1.56 \cdot 500}{0.512} \)

Finally, multiply \( m_{\text{solute}} \) by the molecular weight of the substance (50.0 g/mol) to get the mass of the substance required in grams:

\( \text{mass of substance} = m_{\text{solute}} \cdot \text{molecular weight} \)

To find the mass of the molecular substance that needs to be added to the water, we need to use the equation:

\( \Delta T = K_b \cdot m \)

where:
- \(\Delta T\) is the boiling point elevation (the difference between the boiling point of the solution and the boiling point of the pure solvent),
- \(K_b\) is the molal boiling-point elevation constant of the solvent,
- \(m\) is the molality of the solute (molecular substance) in the solution.

In this case, the boiling point elevation is given as \(101.56\) °C. We first convert this to Kelvin:

\( \Delta T(K) = 101.56 \, \text{°C} + 273.15 \)

Next, we need to find the molality of the solute. To do this, we use the formula:

\( m = \frac{n}{m_{\text{solvent}}} \)

where:
- \( n \) is the number of moles of the solute,
- \( m_{\text{solvent}} \) is the mass of the solvent in kilograms.

We know the mole weight of the molecular substance is \( 50.0 \, \text{g/mol} \). The moles (\( n \)) can be obtained by dividing the mass in grams by the molar weight of the substance:

\( n = \frac{\text{mass in grams}}{\text{molar weight}} \)

Given that the mass of water is \( 500 \, \text{g} \), we convert this to kilograms (\( m_{\text{solvent}} \)) by dividing by \( 1000 \).

Finally, we can substitute the known values into the equation:

\( \Delta T(K) = K_b \cdot m_{\text{solvent}} \)

and solve for the mass of the molecular substance.

delta T = Kb*m

solve for m

m = moles/kg solvent
solve for moles.

moles = grams/molar mass.
solve for grams.

Post your work if you get stuck.

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