To 100 ml CaCl2 solution containing 6.66g CaCl2, 100 ml of 0.5M Na3PO4 solution is added. The mass of Ca3(PO4)2 precipitated would be ?
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Answer:
amount of CaCl2=6.6g
moles of Na3Po4 =0.5 M
now we have to find molar mass of Na3Po4=G.M/M.M
=226/0.5
=4.52g.
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@ ANUSHA ❤✌
=4.52/6.6
=6.2g
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