The volume (in `mL`) of `0.1M Ag NO_(3)` required for complete precipitation of chloride ions present in `30 mL` of `0.01M` solution of `[Cr(H_(2)O)_(5)Cl]Cl_(2)`, as silver chloride is close to:
Text Solution
Verified by Experts
`30 " mL of " 0.01 M [Cr(H_(2)O)_(5)Cl]Cl_(2)implies2Cl^(ɵ)` ions `therefore` Number of `Cl^(ɵ)` ions`=30xx0.01xx2=0.6mol` `therefore` mmoles of `Ag^(o+)` ions required `=0.6 mmol` `therefore` volume `=?` `Vxx0.1=0.6impliesV=6mL`
The volume (in mL) of 1.0 M AgNO_3 required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H_2O)_6]Cl_3 , as silver chloride is close to
Volume (in ml) of 0.7 M NaOH required for complete reaction with 350 ml of o.3 M H_(3) PO_(3) solution is
The number of Cl^(-) ions in 100 mL of 0.001 M HCl solution is
Knowledge Check
Number of chloride ions in 100 mL of 0.01 M HCl solution are
A
`6.023 xx 10^(22)`
B
`6.023 xx 10^(21)`
C
`6.023 xx 10^(20)`
D
`6.023 xx 10^(24)`
The volume (in ml) of 0.5 M NaOH required for the complete reaction with 150 ml of 1.5M H_(3)PO_(3) solution is
A
1350
B
900
C
1250
D
1150
Volume (in ml) of 0.7 M NaOH required for complete reaction with 350 ml of o.3 M H_(3) PO_(3) solution is
