20.00 cm\(^3\) of a solution containing 0.53 g of anhydrous Na\(_2\)CO\(_3\) in 100 cm\(^3\) requires 25.00 cm\(^3\) of H\(_2\)SO\(_4\) for complete neutralisation. The concentration of the acid solution in moles per dm\(^3\) is [H =1, C = 12, O = 16, Na =23, S = 32]
0.02
0.04
0.06
0.08
Explanation
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Discussions (22)
it pretty simple
n=C×V/1000
n=m/M
m/M=C×V/1000
C=m×1000/M×V
C=0.53×1000/106×100
C=530/10600=0.05M
H2SO4+Na2CO3----------- Na2SO4+CO2+H2O
CaVa/CbVb=na/nb
Ca×25/20×0.05=1/1
Ca=20×0.05×1/25×1
Ca=1/25=0.04M or moldm-³
conc. in moles/dm³=molar conc=mole
100cm³ of sol.contain 0.53gof NaCO3
100cm of solution will contain 0.53×1000ç/100cm³
molarity=concentration/molar mass =5.3/106=5.3
EQUATION OF THE REACTION
NaCO3+H2SO4>>>>>>NASO4+CO2+H2O
for molarity of acid ,we have Ma×Va=MB×VB
MA=MB×VB/VA
=0.05×20/25= 0.04 answer (B)
Solution:
C1v1=C2V2
C1V1=n1
n1=m/M
m/M =C2v2
0.56/106=C2v2
0.00528=C2V2
0.0052830=(C2)(25)
C2=0.0052830÷25 =0.000211
Since it was combined in 100cm3 you will multiply by 100
C2=0.000211×100
C2=0.02
Following this i got A as the answer
Note:Disregard this solving as it is obviously not correct
I just wanted to show you guys how i arrived at an incorrect answer
Bros nnaeh Copy questions wella now! Where is de Molarity of acid! And. The question is not even complete at that
