Calculate the molarity of a solution obtained by mixing 100ml of 0.3M H2SO4 and 200ml of 1.5M H2SO4
Answers
Answered by
14
By using the equation
we get,
0.3*100 + 1.5*200 = 330M
we get,
0.3*100 + 1.5*200 = 330M
Answered by
26
here, we using concept , of mixture .
MV = M1V1 + M2V2 + M3V3 + .......
where,
M and V are the final molarity and volume of solution respectively .
M1 , M2 , M3 , M4........ are molatity of components of mixing substances and V1, V2, V3 , .......are the volume of components of mixing substance In solution.
here, we will take ,
MV = M1V1 + M2V2
where,
M1 = 0.3M
V1 = 100 mL
M2 = 1.5 M
V2 =200mL
V= V1 + V2 = 100 + 200 = 300mL
M = ?
then,
M× 300 = 0.3 × 100 + 1.5 × 200
M × 300 = 30 + 300 = 330
M = 330/300 = 1.1 M
hence, molarity of solution = 1.1M
MV = M1V1 + M2V2 + M3V3 + .......
where,
M and V are the final molarity and volume of solution respectively .
M1 , M2 , M3 , M4........ are molatity of components of mixing substances and V1, V2, V3 , .......are the volume of components of mixing substance In solution.
here, we will take ,
MV = M1V1 + M2V2
where,
M1 = 0.3M
V1 = 100 mL
M2 = 1.5 M
V2 =200mL
V= V1 + V2 = 100 + 200 = 300mL
M = ?
then,
M× 300 = 0.3 × 100 + 1.5 × 200
M × 300 = 30 + 300 = 330
M = 330/300 = 1.1 M
hence, molarity of solution = 1.1M
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