A water sample contains 9.5% MgCl2 and 11.7% NaCl by weight. Assuming 80% ionization of each salt, boiling point of water will be approximately?
(kb=0.52)
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Answer:
Explanation:
Solution :
Moles of Nacl in 100 g of solution =11.7/54=0.2
Similarly moles of MgCl2=9.5/95=0.1
weight of solvent =100−(11.7+9.5)=78.8g
no of moles of ions from Nacl(80%ioniz))=0.36
no of moles of ions from Mgcl2(80%ioniz))=0.36
(n1+n2)i=0.36+0.2=0.56
=1000×0.52×0.56/78.86=3.69
=100+3.69=103.69edges
sheemaf67:
In the above answer, how to calculate no.of moles of ion from NaCl(80% Ionia)
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