Question

In 50 mL of 0.5 M H2SO4, 75 mL of 0.25 M H2SO4 is added. What is the concentration of the final solution if its volume is 125 mL?

its urgent...

Answer 1

$\textbf{ Solution }$

No. of moles in 0.05 l of $H_2 SO_4$

= M × V(in liter)

= 0.5 × 0.05

= 0.025

No. of moles in 0.075 l of $H_2 SO_4$ that are added

$\implies 0.25 \times 0.075}$

$\implies 0.01875}$

Now,

Total no. of moles in 0.125 l of $H_2 SO_4$

$\implies 0.025 + 0.01875 }$

$\implies 0.04375 }$

∴ Molarity of H2SO4 = $\implies \frac{ 0.04375 }{0.0125}$

$\implies = 0.35 M. }$

Answer 2

Heya mate
The answer of ur question is



♢ final concentration=M1v1+M2V2/total volume

M1=0.5M
V1=50ml/1000 = 0.05L
M2=0.25M
V2=150ml/1000
= 0.15L

final volume = 250ml(given in the question (made to 250ml))
= 250ml/1000
= 0.25

by applying the formula ,we get that

final concentration = 0.05*0.5+0.15*0.25/0.25
so the answer is 0.25M

final concentration will be 0.25M




hope it helps