Question

Two solutions of H2SO4 are prepared with different composition.
Solution A: 500 mL of 1 M aq. H2SO4 having density 1.2 g/mL.
Solution B: 100 mL of 2 M aq. H2SO4 having density 1.4 g/mL.
Select the correct statement.
If both solutions A and B are mixed, then molarity of final solution is 2.33 M
If both solutions A and B are mixed, then (w/w)% of H2SO4 is 10.72% in final solution
If both solutions A and B are mixed, then (w/v)% of H2SO4 is 11.43% in final solution
If both solutions A and B are mixed, then normality of final solution is 1.8 N​

1​

Answer 1

All the statements are wrong.

Given,

Volume of solution-A=500 mL

Molarity of solution-A=1 M

Density of solution-A=1.2 g/mL

Volume of solution-B=100 mL

Molarity of solution-B=2 M

Density of solution-B=1.4 g/mL.

To find,

the correct statement among the statements given in the question.

Solution:

• Molarity of a substance in a solution is defined as the moles of that substance in 1 L of solution.
• $Molarity=\frac{Moles}{Volume(L)}$.
• Molarity is a temperature dependent concentration term.
• When two solutions are mixed, the molarity of the resultant solution is found by the following equation:
• $M1V1+M2V2=MfVf$.

The molarity of the resultant solution will be,

$M1V1+M2V2=MfVf\\1X\frac{500}{1000} +2X\frac{100}{1000}=MfX\frac{500+100}{1000} \\1X\frac{5}{10} +2X\frac{1}{10}=MfX\frac{600}{1000} \\\frac{5}{10} +\frac{2}{10}=MfX\frac{6}{10}\\\frac{5+2}{10} =MfX\frac{6}{10}\\\frac{7}{10} =MfX\frac{6}{10}\\Mf=\frac{7X10}{6X10} M\\Mf=\frac{7}{6} M\\Mf=1.17 M.$

The molarity of the resultant solution is 1.67 M, so the first statement is wrong.

The weight of solution-A is,

$Density=\frac{Mass}{Volume} \\1.2=\frac{Mass}{500} \\Mass=1.2X500\\Mass=600 g.$

The weight of solution-B is,

$Density=\frac{Mass}{Volume} \\1.4=\frac{Mass}{100} \\Mass=1.4X100\\Mass=140 g.$

The weight of the resultant solution will be,

$=(600+140)g\\=740 g.$

Moles of H2SO4 in solution-A,

$Molarity=\frac{Moles}{Volume(L)} \\1=\frac{Moles}{\frac{500}{1000} } \\1=\frac{Moles}{0.5} \\Moles=0.5X1\\Moles=0.5.$

The molar mass of H2SO4 is 98 g.

The mass of 0.5 moles of H2SO4 in solution-A,

$Moles=\frac{Mass}{Molar mass} \\0.5=\frac{Mass}{98} \\Mass=0.5X98\\Mass=49 g.$

Moles of H2SO4 in solution-B,

$Molarity=\frac{Moles}{Volume(L)} \\2=\frac{Moles}{\frac{100}{1000} } \\2=\frac{Moles}{0.5} \\Moles=0.5X2\\Moles=1.$

As 1 mole is present in solution-B, its mass will be equal to its molar mass i.e 98 g.

Total mass of H2SO4 in the final solution,

$=(98+49)g\\=147 g.$

The (w/w)% of the solution will be,

(w/w)%=(Mass of solute/Mass of solution) x 100

$w/w=\frac{147}{740} X100\\w/w=19.86 percent.$

The (w/w)% of the solution is 19.86%, thus the second statement is also wrong.

Total volume of the final solution,

$=(500+100)mL\\=600mL.$

Total mass of H2SO4 in the final solution,

$=(98+49)g\\=147 g.$

The (w/v)% of the solution will be,

(w/v)%=(Mass of solute/Volume of solution) x 100

$w/v=\frac{147}{600} X 100\\w/v=\frac{147}{6}\\w/v=24.5 percent.$

The (w/v)% of the solution is 24.5%, thus the third statement is also wrong.

The n-factor of H2SO4 is 2.

$Normality=MolarityXn-factor.$

The normality of solution-A will be,

$=1X2\\=2 N.$

The normality of solution-B will be,

$=2X2\\=4 N.$

The normality of the final solution is given by,

$Nf=\frac{N1V1-N2V2}{V1+V2}.$

where, $N1V1 > N2V2.$

The normality of the final solution will be:

$Nf=\frac{2X500-4X100}{500+100} \\Nf=\frac{1000-400}{600} \\Nf=\frac{600}{600} N\\Nf=1N.$

The normality of the final solution is 1 N, so the fourth statement is also wrong.

Hence, all the statement are wrong.

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