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3 years ago

How many molecular of H2O and O2 are present in 8.5g of H2O2 ?

Chemistry

1 answer:

Mama L [17]3 years ago

4 0

2H2O+O2--->2H2O2

8.5 gm H2O2=0.25 mole

hence H2O is also 0.25 mole i.e.4.5 gm

O2is 0.125 mole i.e.4 gm

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Brass is a substitutional alloy consisting of a solution of copper and zinc. A particular sample of red brass consisting of 79.0

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Answer: The molality and molarity of zinc in the solution is 4.07 m and 28.11 M respectively.

Explanation:

Solute is the substance which is present in smaller proportion in a mixture and solvent is the substance which is present in larger proportion in a mixture.

We are given:

(m/m) % of Cu = 79 %

This means that 79 g of copper is present in 100 grams of alloy.

(m/m) % of Zn = 21 %

This means that 21 g of zinc is present in 100 grams of alloy.

As, zinc is present in smaller proportion. So, it is solute and copper is the solvent.

Calculating molality of zinc:

To calculate molality of the zinc, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$m_{solute}$ = Given mass of solute (Zinc ) = 21 g

$M_{solute}$ = Molar mass of solute (Zinc) = 65.3 g/mol

$W_{solvent}$ = Mass of solvent (copper) = 79 g

Putting values in above equation, we get:

$\text{Molality of Zinc}=\frac{21\times 1000}{65.3\times 79}\\\\\text{Molality of zinc}=4.07m$

Calculating molarity of zinc:

To calculate volume of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = $8740kg/m^3=\frac{8740kg\frac{1000g}{1kg}}{1m^3\times \frac{10^6cm^3}{1m^3}}=\frac{8740g}{1000cm^3}=8.740g/cm^3$

(Conversion factors used are: 1 kg = 1000 g & $1m^3=10^6cm^3$ )

Mass of solution = 100 g

Putting values in above equation, we get:

$8.740g/cm^3=\frac{100.0g}{\text{Volume of zinc}}\\\\\text{Volume of zinc}=11.44cm^3$

To calculate the molarity of solution, we use the equation:

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

We are given:

Molar mass of zinc = 65.3 g/mol

Volume of solution = $11.44cm^3=11.44mL$ (Conversion factor: $1cm^3=1mL$ )

Mass of zinc = 21.0 g

Putting values in above equation, we get:

$\text{Molarity of zinc}=\frac{21\times 1000}{65.3\times 11.44}=28.11M$

Hence, the molality and molarity of zinc in the solution is 4.07 m and 28.11 M respectively.

4 0

3 years ago

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3 years ago

How many molecular of H2O and O2 are present in 8.5g of H2O2 ?​

How many molecular of H2O and O2 are present in 8.5g of H2O2 ?