Chemistry, 09.01.2020 06:31 nevelle

A5.05% (by mass) of aqueaus solution of sodium sulfate
isallowed to react with an excess of barium chloride. what volume
ofthe sodium sulfate solution (density= 1.03g/ml) is required
toproduce 2.73 g of barium

Answers: 3

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Answer from: ToonGamesToo

Answer : The volume of the sodium sulfate solution required is, 0.0319 L

Explanation :

As we are given that 5.05 % (by mass) of aqueous solution of sodium sulfate that means, 5.05 grams of sodium sulfate present in 100 grams of solution.

Mass of sodium sulfate = 5.05 g

Mass of solution = 100 g

Density of solution = 1.06 g/mL

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\farc{100g}{1.03g/mL}=97.09mL$

Now we have to calculate the molarity of sodium sulfate solution.

Formula used :

$\text{Molarity of sodium sulfate solution}=\frac{\text{Mass of sodium sulfate}\times 1000}{\text{Molar mass of sodium sulfate}\times \text{Volume of solution (in mL)}}$

Molar mass of sodium sulfate = 142.04 g/mol

Now put all the given values in this formula, we get:

$\text{Molarity of sodium sulfate solution}=\frac{5.05g\times 1000}{142.04g/mole\times 97.09mL}=0.366mole/L$

Now we have to calculate the moles of barium sulfate.

$\text{Moles of }BaSO_4=\frac{\text{Mass of }BaSO_4}{\text{Molar mass of }BaSO_4}$

Molar mass of barium sulfate = 233.38 g/mol

$\text{Moles of }BaSO_4=\frac{2.73g}{233.38g/mol}=0.0117mol$

Now we have to calculate the moles of sodium sulfate.

The balanced chemical reaction will be:

$Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4$

From the balanced chemical reaction, we conclude that

As, 1 mole of BaSO_4 produced from 1 mole of Na_2SO_4

So , 0.0117 mole of BaSO_4 produced from 0.0117 mole of Na_2SO_4

Now we have to calculate the volume of sodium sulfate.

$Molarity=\frac{Moles}{Volume}$

$0.366mol/L=\frac{0.0117mol}{Volume}$

Volume=0.0319L

Thus, the volume of the sodium sulfate solution required is, 0.0319 L

Answer from: Quest

the correct answer is boyle

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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A5.05% (by mass) of aqueaus solution of sodium sulfate
isallowed to react with an excess of ba...