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WHAT WILL BE THE CONCENTRATION OF CHLORIDE ION IF 200 ml of 0.2 M CaCl2 IS MIXEDWITH 300 ml OF 0.1 MNaCl SOLUTION ?

RAGINI SEIWAL , 6 Years ago
Grade 12
anser 1 Answers
mrunali

Last Activity: 6 Years ago

HERE IS THE DIRECT FORMULA TO CALCULATE THE RESULTANT CONCENTRATION-
M(V1 + V2) = M1V1 + M2V2
HERE M1 IS THE CONCENTRATION OF CHLORIDE FROM CaCl2 = 2 * 0.2 (as in CaCl2 2 Cl are present ) and V1=200
M2 IS THE CONCENTRATION OF CHLORIDE FROM Nacl = 0.1 AND V2 =300
SO AFTER PUTTING VALUES THE ANSWER WILL BE 
M* 500 = 200*0.4 + 300*0.1
M=110/500
M= 0.22

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