Question

100 ml of 49% (by mass) H2SO4 (d=1.5g/ml) is mixed with 100 ml 2.5M H2SO4 .find the molarity of the solution will be

Answer 1

Answer : The molarity of the solution will be, 3.75 mole/L

Solution :  Given,

49 % by mass means that 49 gram of $H_2SO_4$ present in 100 gram of solution.

Molar mass of $H_2SO_4$ = 98 g/mole

First we have to calculate the molarity of $H_2SO_4$.

$Molarity=\frac{\text{Mass of }H_2SO_4\times 1000}{\text{Molar mass of }H_2SO_4\times \text{Volume of solution}}=\frac{49g\times 1000}{98g/mole\times 100ml}=5mole/L=5M$

Now we have to calculate the molarity of the solution.

According to the dilution law,

$M_1V_1+M_2V_2=MV$

where,

$M_1$ = molarity of $H_2SO_4$ solution = 5 M

$V_1$ = volume of $H_2SO_4$ solution = 100 ml = 0.1 L

$M_2$ = molarity of another $H_2SO_4$ solution = 2.5 M

$V_2$ = volume of another $H_2SO_4$ solution = 100 ml = 0.1 L

M = molarity of solution = ?

Volume of solution = 0.1 + 0.1 = 0.2 L

Now put all the given values in the above law, we get the molarity of solution.

$(5M)\times (0.1L)+(2.5M)\times (0.1L)=M\times (0.2L)$

$M=3.75mole/L=3.75M$

Therefore, the molarity of the solution will be, 3.75 mole/L