calculate the molarity of 30ml of 0.5M of H2SO4 diluted to 500ml
Answers
the molarity of 30ml of 0.5M of H2SO4 diluted to 500ml
*EXPLANATION*
Volume of H2SO4 (v 1) = 30ml
diluted volume of H2SO4 (V2) = 500ml
number of moles present in 1000ml of 0.5 m H2SO4 = 0.5 mol
Intial molarity M1 = 0.5m
We have to find M2 ( molarity of 500ml solution )
M2V2 = M1V1
M2 × 500 = 0.5 × 30
M2 = 15 / 500 = 0.03 m
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answer
method 1 :
using the formula M1V1 = M2V2 (dilution equation) where M1 is molarity of original solution , V1 is volume of original solution , M2 is molarity of diluted solution and V2 is volume of diluted solution.
initially we have, M1 = 0.5 mol , V1 = 30 mL
after diluting we have, V2 = 500mL
Now, value of M2 is to be found
we know that, M2V2 = M1V1
=> M2 = M1V1/V2
(putting values)
=> M2 = (0.5×30)/500 = 15/500
=> M2 = 0.03M
therefore, molarity is 0.03M.
method 2 :
molarity = no.of moles of solute ÷ volume of solution (in litres)
given, volume of soln. = 500mL
we have to find no.of moles.
given, 0.5M of H2SO4.
∴ no.of moles of H2SO4 in 1L or 1000mL = 0.5mol
=> no.of moles in 1 mL = 0.5/1000 mol
=> no.of moles of H2SO4 in 30ml of 0.5M of H2SO4 = 30 × (0.5/1000) = 15/1000 = 0.015 mol
∴ molarity = 0.015mol/500mL = 0.03M