100ml of h2s04 solution having molarity 1M and density 1.5g/ml is mixed with 400ml of water. Calculate final molarity of h2s04 solution if final density 1.25g/ml.

Question

Ans is 0.227M
Who help me thanks a lot

Answer 1

I never worked a problem like this before.
mass 100 mL of 1M H2SO4 = 1.5 g/mL x 100 mL = 150 grams.
mass H2SO4 in that 100 mL is 98 g/mol x 100/1000 = 9.8 g H2SO4
g H2O in that 100 solution is 150g - 9.8 g H2SO4 = 140.2

g H2O in the final solution is 400 + 140.2 = 540.2 and final mass of that solution is 540.2 + 9.8 g H2SO4 = 550 grams.
Then 1.25 g/mL x #mL = 550 g
# mL solution is 440 mL.
Then mL1 x M1 = mL2 x M2
100 x 1 = 440 x M2
Solve for M2 = 0.227

Answer 2

0.227m

Answer 3

no of moles = molarity x vol
no of moles of H2SO4 = 1 x 100/1000 = 0.1

final vol = total weight /density
( wt of H2SO4 + wt of H20)/ 1.25
(1.5 x 100 + 400 x1 ) 1.25
= 550/1.25 = 440 ml

molarity of final solution = moles/vol in litre
0.1 /440/1000 = 0.227