1. Assuming the volumes are additive; i.e., 100 + 400 = 500 mL, then
M H2SO4 = 1 M x (100/500) = ?
2.
I don't know how to deal with 938% H2SO4; however, that never enters into the problem. I assume that 20.8% BaCl2 is w/v and that means 20.8 g BaCl2/100 mL. Therefore, in 50 mL there must be just half that or 10.4g BaCl2.
mols BaCl2 in 10.4g BaCl2 = 10.4g/molar mass BaCl2 = approx 0.05. mols Cl^- is twice that = or approx 0.1 mol and that in 50 mL is M = mols/L = 0.1 mol/0.05L = approx 2 M.
That is diluted from 50 to 150 mL so 2M x (50/150 mL) = approx ?
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 is 1.25g/ml.
50ml of 20.8% bacl2 and 100ml of 938% h2s04 solutions are mixed. Molarity of cl ions in the resulting solutions is at. wt of ba =137
Who helps me solve two questions
Thank a lot
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