Question

A solution of H2O2   is labelled as 11.2 V.  If the density of solution is 1.034 g/mL then identity the correct option.(Multiple choice answer)​

Answer 1

Answer:

The answer is a option C

Answer 2

Answer:

The molality and % weight upon the volume of the solution is $\mathrm{\frac{1}{1.034} }$ and $\mathrm{\frac{34}{1000} }$ respectively.

Explanation:

First, let's calculate the molarity of the solution,

$\mathrm{M=\frac{v}{11.2} }$         (1)

Where,

M=molarity

v=volume strength of the solution

From the question we have,

v=11.2

The density of the solution(ρ)=1.034 g/mL

By inserting the value of "v" in equation (1) we get;

$\mathrm{M=\frac{11.2}{11.2} }$

$\mathrm{M=1}$             (2)

From the value of molarity, we can deduce that 1 liter contains 1 mole of H₂O₂.

Mass of H₂O₂=1×34=34 gram    (Molar mass of H₂O₂=34 gram)

Weight of one liter solution=1.034 g/mL=1.034×10³ g/L=1034 g/L

$\mathrm{W_{H_2O}=1034-34=1000\ grams}$          (3)

Moles of H₂O=$\frac{1000}{18}=55.56$             (4)

Now we need to calculate the molality,

Molality 'm'=$1\times \frac{1000}{1034}\ \mathrm{or}\ \frac{1}{1.034} =0.96$           (5)

Mole fraction,

$\mathrm{x_{H_2O_2}=\frac{n_{H_2O_2}}{n_T} }$

$\mathrm{x_{H_2O_2}=\frac{1}{56.56} }$

$\mathrm{x_{H_2O_2}=0.01768 }$                           (6)

Now, let's find the % weight by volume of the solution,

$\mathrm{\%\frac{w}{v}=\frac{1\times 34}{1000} =\frac{34}{1000} }$             (7)

By analyzing equations (5) and (7) we can say that the correct answers are (b) and (d).

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