1.

What is the mass of the precipitate formed when 50 mL of 17.0% solution of AgNO_(3) is mixed with 50 mL of 11.6%NaCl solution?

Answer»

Solution :`" 100 ML of AgNO"_(3)" solution contain AgNO"_(3)=17.0 g`
`THEREFORE"50 mL of AgNO"_(3)" solution will contain AgNO"_(3)=(17.0)/(100)xx50=8.5g`
`"Molar mass of AgNO"_(3)=108+14+3xx16=170" g mol"^(-1)`
`therefore"Moles of AgNO"_(3)" present"=(8.5)/(170)="0.05 mol"`
`"100 mL of NACL solution contain NaCl = 11.6 g"`
`therefore"50 mL of NaCl solution will contain NaCl"=(11.6)/(100)xx50=5.8g`
`"Molar mass of NaCl"=23+35.5=58.5" g mol"^(-1)`
`therefore"Moles of NaCl present "=(58)/(8.5)=" 0.1 mole"`
The reaction taking place will be
`{:(,AgNO_(3),+,NaClrarrAgCl+NaNO_(3)),("Amounts taken","0.05 mole",,"0.1 mole"):}`
Thus, `AgNO_(3)` will be the limiting reagent.
`"AgCl formed = 0.05 mole "=0.05xx143.5g=7.175g`


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