Suppose 3.005 g of a nonvolatile solute is added to 20.02 g of water (the solvent), and the boiling point increases from 100.000 OC to 101.286 OC. Determine the TB, molality, moles, and molecular weight for the solute if kb for water is 0.512 OC/m. Report each value using the correct number of significant digits. Refer to Example 1.2 and pages 3-4 in the chapter 1 notes for general chemistry 1 to understand significant figures.

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Explanation:

Hello,

In this case, considering the boiling point rise problem, we consider its appropriate equation:

Whereas i is the van't Hoff factor that for this nonvolatile solute is 1, m is the molality, Kb the boiling point constant of water as it is the solvent and ΔT the temperature difference. In such a way, with the given information we obtain:

- ΔT:

- Molality (mol/kg):

- Moles for 20.02 g (0.02002 kg) of water:

- Molar mass:

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as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2