If 1 litre of 2.2M sulphuric acid is poured into a bucket containing 10 litre of water,and resulting solution mixed thoroughly, the resulting sulpuric acid concentration will be?
2.2 M
1.1 M
O.22 M
0. 11 M
0.20 M
Explanation
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C1V1=C2V2....where v2 =v1+new volume
2.2×1=C2×(1+10)
2.2=C2×11
2.2÷11=C2
C2 =O.2M
Wrong option!
the Value of V1 is that of the acid and not that of water!
V1 = 1litre
The value of V2 = 1litre of sulphuric acid + 1litre of water = 1 + 10 = 11
therefore,
C1V1 = C2V2
2.2 × 1 = C2 × 11
C2 = 0.2M
Molar Conc or Molarity(M) = no of moles of substance / volume of solution (L or dm3)
Lets find the no of moles of sulphuric acid that was poured in 10L of water.
From Molar Conc formula above, no of moles = molar conc * volume = 2.2M x 1L = 2.2 mol
Now lets find the conc of 2.2 mol of sulphuric acid in 10L of water.
Molar Conc = 2.2 mol / 10L = 0.22 M
One lit of water=1000cm cube
therefore 10 lit of water = 10000cm cube
...2.2×1000\10000=0.22
Prof.,Felicia's correct.
To cal.concentration{M},we use M=m/g. Buh remember;cm^3 is equivalent to g.
Hence,M=m/g=2.2/10=0.22moldm^-3.
OR
Using the general formula for finding concentration of a subt.;
M={Reacting mass/Molar mass} * {1000/gram of H2O}.
Buh since we're looking for conc.of the mass of reacting substance where the main concentration has been given,thus;
2.2={g/98 * 1000/10}. 2.2=1000g/980. g=215.6/1000,g=0.2156=0.22{approximately}.
N/B: The velue 98 was got from the mathematical value of Sulphuric acid{H2SO4}.
