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miskamm [114]
3 years ago
14

What is the molality of a solution of water and kcl if the freezing point of the solution is –3mc030-1.jpgc?

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
4 0
We will use the expression for freezing point depression ∆Tf
     ∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is 
     ∆Tf = 0C - (-3°C) = 3°C 
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and  one Cl- ion:
     KCl → K+ + Cl- 
Therefore, the molality m of the solution can be calculated as 
     3 = 2 * 1.86 * m
     m = 3 / (2 * 1.86)
     m = 0.80 molal
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A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
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<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

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How many chlorine atoms are on the products side of the reaction 2al 6hcl → 2alcl3 3h2? 2 3 6 9
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The number of chlorine atoms present on the product side of the reaction is 6

<h3>What is a chemical equation? </h3>

Chemical equations are representations of chemical reactions using symbols and formula of the reactants and products.

The balancing of chemical equations follows the law of conservation of matter which states that matter can neither be created nor destroyed during a chemical reaction but can be transferred from one form to another.

<h3>How to determine the number of atoms of Cl</h3>

2Al + 6HCl → 2AlCl₃ + 3H₂

Products => AlCl₃ and H₂

Number of Cl atoms = 2 × 3

Number of Cl atoms = 6 atoms

Learn more about chemical equation:

brainly.com/question/7181548

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