Chemistry, 05.05.2020 21:15 amanzi10

How many grams of sodium bromide must be dissolved in 400.0 g of water to produce a 0.500 molal solution?

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Answer from: hgiaculliozdjov

The answer is 20.6 grams.

Explanation:

Molality describes the concentration of a solution. It can be defined as the number of moles of a solute dissolved in 1 kilogram of solvent. Then it is equal to the moles of solute (the substance that dissolves) divided by the kilograms of solvent (the substance used to dissolve):

$Molality=\frac{number of moles of solute}{kilogram of solvent}$

The molality is expressed in units ( $\frac{moles}{kg}$ ).

So, you can apply the following rule of three with the solution being 0.5 molal: if in 1 kg of solution there are 0.5 moles of solute, in 0.4 kg (400 g, being 1kg = 1000g) how many moles of solute are there?

$moles=\frac{0.4 kg*0.5 moles}{1 kg}$

moles=0.2 moles

Now, you know:

Na: 23 g/moleBr: 80 g/mole

Then, The molar mass of sodium bromide NaBr is

NaBr= 23 g/mole + 80 g/mole= 103 g/mole

Now a new rule of three applies, if in 1 mole of sodium bromide there are 103 grams, in 0.2 mole how much mass is there?

$mass=\frac{0.2 moles*103 grams}{1 mole}$

mass= 20.6 grams

The answer is 20.6 grams.

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

1. and 2. question: in which equation? or i don't understand you, 3. a), 4. c), 5. c), 6. b), 7. d), 8. d) if small number is underlined(behind symbol) , b) if big number is underlined(in front of symbol)9. what is underlined there? 10. c)

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