Question

Calculate the molarity of a solution obtained by mixing 100ml of 0.3M H2SO4 and 200ml of 1.5M H2SO4

Answer 1

By using the equation 
$M_{3}V_{3}=M_{1}V_{1}+M_{2}V_{2}$
we get,
0.3*100 + 1.5*200 = 330M

Answer 2

here, we using concept , of mixture .
MV = M1V1 + M2V2 + M3V3 + .......
where,
M and V are the final molarity and volume of solution respectively .
M1 , M2 , M3 , M4........ are molatity of components of mixing substances and V1, V2, V3 , .......are the volume of components of mixing substance In solution.

here, we will take ,
MV = M1V1 + M2V2

where,
M1 = 0.3M
V1 = 100 mL
M2 = 1.5 M
V2 =200mL
V= V1 + V2 = 100 + 200 = 300mL
M = ?

then,
M× 300 = 0.3 × 100 + 1.5 × 200

M × 300 = 30 + 300 = 330

M = 330/300 = 1.1 M

hence, molarity of solution = 1.1M