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Answer :
Answer:
The molality of HCl solution is 16.24 mol/kg.
The molality of [tex]HC_2H_3O_2[/tex] solution is 82,500 mol/kg.
The molality of [tex]NH_3[/tex] solution is 27.78 mol/kg.
Explanation:
formula used:
[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]
1) Mass percentage of the HCl solution = 37.2%
This means that in 100 grams of solution 37.2 grams of HCl is present.
Mass of HCl (solute)= 37.2 g
Mass of water(solvent) = 100 g - 37.2 g = 62.8 g = 0.0628 kg (1g = 0.001 kg)
Mole of HCl = [tex]\frac{37.2 g}{36.465 g/mol}=1.020 mol[/tex]
[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]
[tex]m=\frac{1.020 mol}{0.0628 kg}=16.24 mol/kg[/tex]
The molality of HCl solution is 16.24 mol/kg.
2) Mass percentage of the [tex]HC_2H_3O_2[/tex] solution = 99,8%
This means that in 100 grams of solution 99.8 grams of [tex]HC_2H_3O_2[/tex] is present.
Mass of [tex]HC_2H_3O_2[/tex](solute)= 99.8 g
Mass of water(solvent) = 100 g - 99.8 g = 0.2 g = 0.0002 kg (1g = 0.001 kg)
Mole of [tex]HC_2H_3O_2[/tex] = [tex]\frac{99.8 g}{60.05g/mol}=16.50 mol[/tex]
[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]
[tex]m=\frac{16.50 mol}{0.0002 kg}=82,500 mol/kg[/tex]
The molality of [tex]HC_2H_3O_2[/tex] solution is 82,500 mol/kg.
3) Mass percentage of the [tex]NH_3[/tex] solution = 28.0%
This means that in 100 grams of solution 28.0 grams of [tex]NH_3[/tex] is present.
Mass of [tex]NH_3[/tex](solute)= 37.2 g
Mass of water(solvent) = 100 g - 28.0 g = 72.0 g = 0.072 kg (1g = 0.001 kg)
Mole of [tex]NH_3[/tex]= [tex]\frac{28.0g}{17 g/mol}=2mol[/tex]
[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]
[tex]m=\frac{2 mol}{0.072kg}=27.78 mol/kg[/tex]
The molality of [tex]NH_3[/tex] solution is 27.78 mol/kg.
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Final answer:
To calculate the molalities of HCl, HC2H3O2, and NH3, convert the weight percentage to mass of solute, then convert that to moles using the formula weight, and divide by the mass of the solvent in kilograms. Use the provided densities to infer the mass of water solvent and proceed with calculations respecting significant figures.
Explanation:
Calculating Molalities of Commercial Reagents
To calculate the molalities of commercial reagents, you'll first need to determine the moles of solute and then divide that by the mass of the solvent in kilograms. Below is the process for each reagent provided:
For HCl (hydrochloric acid), we use the weight % to find the mass of HCl per 100g of solution, convert that mass to moles using the formula weight, and then divide by the solvent mass (water) in kilograms to find the molality.
In the case of HC2H3O2 (acetic acid), we apply a similar approach, starting with the weight % to calculate the mass of acetic acid in 100g of solution, convert to moles using its formula weight, and then divide by the mass of solvent to find molality.
For NH3 (aqueous ammonia), we begin by taking the 28.0 weight % value to get the mass of NH3 in 100g of solution, converting this mass to moles using NH3's formula weight, and finding the molality by dividing moles over kilograms of solvent.
Molality (°m) is defined as the number of moles of solute per kilogram of solvent and is especially useful because it doesn’t change with temperature. Using the provided densities and weights, we can assume the solvent used is predominantly water and calculate the weight of the water to use in the molality equation. The molar mass information provided, such as the formula weight of HCl (36.465 g/mol), acts as a conversion factor between grams and moles of solute.
At each step, remember to account for significant figures based on the precision of the given data.
