a solution of 10.00g of a compound dissolved in 0.200kg of benzene boils at 81.2...

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October 2022 1 1 Report

a solution of 10.00g of a compound dissolved in 0.200kg of benzene boils at 81.2 degree celsius.calculate the molecular weight of the compund pkisagot po ng ayos nonsense answer will be reported thanks po

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LaplaceTransform

✏️BOILING-POINT ELEVATION

$\tt{\huge{\green{Solution:}}}$

The problem asks to solve for the molecular weight of the compound. First, we need to identify the quantities to be used. These are

m = molality

mass = mass of solute

n = number of moles of solute

MW = molecular weight of solute

$\text{kg}_{solv} =$ kilograms of solvent

$K_{\text{b}} =$ molal boiling-point elevation constant

$\Delta T_{\text{b}} =$ boiling-point elevation

$T{{\circ}\atop{\text{b}}} =$ boiling point of the pure solvent

$T_{\text{b}} =$ boiling point of the solution

Second, we need to identify the given values. These are

mass = 10.00 g

$\text{kg}_{solv} =$ 0.200 kg

$K_{\text{b}} =$ 2.53 °C/m

$T{{\circ}\atop{\text{b}}} =$ 80.1°C

$T_{\text{b}} =$ 81.2°C

Third, we need to find the boiling-point elevation of the solution. The boiling-point elevation is

$\Delta T_{\text{b}} = T_{\text{b}} - T{{\circ}\atop{\text{b}}}$

$\Delta T_{\text{b}} = 81.2^{\circ}\text{C} - 80.1^{\circ}\text{C}$

$\Delta T_{\text{b}} = 1.1^{\circ}\text{C}$

Fourth, we need to find the molality of the solution. The molality is

$m = \dfrac{\Delta T_{\text{b}}}{K_{\text{b}}}$

$m = \dfrac{1.1^{\circ}\text{C}}{2.53^{\circ}\text{C/}m}$

m = 0.434783 m

Fifth, we need to find the number of moles of solute. The number of moles of solute is

$n = m(\text{kg}_{solv})$

$n = (0.434783 \: m)(\text{0.200 kg})$

n = 0.0869566 mol

Finally, we can now solve for the molecular weight of solute. Therefore, the molecular weight of solute is

$\text{MW} = \dfrac{\text{mass}}{n}$

$\text{MW} = \dfrac{\text{10.00 g}}{\text{0.0869566 mol}}$

$\boxed{\text{MW = 115 g/mol}}$

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✏️BOILING-POINT ELEVATION

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