Question

3.5 L of 2M HCl solution is mixed with another 3L of 0.5 M HCl solution. Then the molarity of resulting solution is:
i)2.5 M
ii) 1.3 M
iii)0.75 M
iv)None of these
​

Answer 1

ANSWER:

• The molarity of resulting solution = 1.3 M.

GIVEN:

• 3.5 L of 2 M HCl solution is mixed with 3 L of 0.5 M HCl solution.

TO FIND:

• The molarity of resulting solution.

EXPLANATION:

Both the solution are of acidic nature.

$\boxed{\large{\bold{M=\dfrac{M_1V_1+M_2V_2}{V_1+V_2}}}}$

$\sf M_1 = 2\ M$

$\sf V_1 = 3.5\ L$

$\sf M_2 = 0.5\ M$

$\sf V_2 = 3\ L$

$\sf M=\dfrac{2(3.5)+(0.5)3}{3.5+3}$

$\sf M=\dfrac{7+1.5}{6.5}$

$\sf M=\dfrac{8.5}{6.5}$

$\sf Molarity=1.3 \ M$

HENCE THE MOLARITY OF THE RESULTING SOLUTION = 1.3 M.

EXTRA FORMULAE:

★ When solutions are of different nature:

$\boxed{\large{\bold{M=\dfrac{M_1V_1-M_2V_2}{V_1+V_2}}}}$

$\sf Here \ M_1V_1>M_2V_2$

★ Molarity of dilution:

$\boxed{\large{\bold{M_1V_1=M_2V_2}}}$

NORMALITY FORMULAE:

★ When solutions are of different nature:

$\boxed{\large{\bold{N=\dfrac{N_1V_1-N_2V_2}{V_1+V_2}}}}$

$\sf Here \ N_1V_1>N_2V_2$

★ When solutions are of same nature:

$\boxed{\large{\bold{N=\dfrac{N_1V_1+N_2V_2}{V_1+V_2}}}}$

★ Normality of dilution:

$\boxed{\large{\bold{N_1V_1=N_2V_2}}}$