Question

200ml of 1M H2SO4 ,300ml 3M HCL and 100 ml of 2M HCLare mixedand made upto 1litre. The molarity of the solution

Answer 1

Answer : The molarity of the solution is, 1.3 M

Solution :

Formula used :

$M_1V_1+M_2V_2+M_3V_3=MV$

where,

$M_1$ = molarity of $H_2SO_4$ solution = 1 M

$V_1$ = volume of $H_2SO_4$  solution = 200 ml = 0.2 L

$M_2$ = molarity of HCl solution = 3 M

$V_2$ = volume of HCl solution = 300 ml = 0.3 L

$M_3$ = molarity of another HCl solution = 2 M

$V_3$ = volume of another HCl solution = 100 ml = 0.1 L

$M$ = molarity of solution = ?

$V$ = volume of solution = 1 L

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

$1\times 0.2+3\times 0.3+2\times 0.1=M\times 1$

$M=1.3M$

Therefore, molarity of the solution is, 1.3 M

Answer 2

n1v1+n2v2+n3v3/v (total)=(h+)=N(total)

m1v1n1+m2v2n2+m3v3n3/v total =(h+)

1×0.2×2+3×0.3×1+2×0.1×1/1 =(h+)

0.4+0.9+0.2/1=(h+)

1.5 =(h+)

hope u will understand it !!