IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

aspiringiitan17

how many grams of BaCl₂.2H₂O (244 gram per mole) should be taken to prepare 500mL of 0.074M Cl^- solution?

bhuvana89

is the ans 4.514 g ?????

catch_arnnie

solution::

moles of Cl^- prepared = 0.072/2 = 0.036 moles

1 mole BaCl₂.2H₂O gives 2 moles of Cl^-

=> 1 moles of Cl^- will be given by 1/2 mole of BaCl₂.2H₂O

=> 0.036 moles of Cl^- will be given by 0.036/2 = 0.018 moles of BaCl₂.2H₂O

=> mass of BaCl₂.2H₂O required = 244 * 0.018 = 4.392g (ans.)

aspiringiitan17

yes the answer is 4.514g

but hw is it calculated?

aditya928

it will be done like this....

molarity = (no. of moles)/(Volume of Solution in Litrs)

0.074 = (No.of Moles)/0.5L

No. of Moles = 0.074*0.5

No. of Moles = 0.037

(Given mass)/(Mol. Mass) = 0.037

Given Mass = 0.037*244

Given Mass = 9.028/2 ('coz, we want Cl^- not Cl₂)

Given mass = 4.514 gms

catch_arnnie

i just made the mistake in takin 0.072 M instead of 0.074 M.....