Home
Class 12
CHEMISTRY
A sample of NaClO(3) is converted by he...

A sample of `NaClO_(3)` is converted by heat to `NaCl` with a loss of `0.16g`of oxygen. The residue is dissolved in water and precipitated as `AgCl`. The mass of `AgCl` (in g) obtained will be: (Given : Molar mass of `AgCl = 143.5 g mol^(-1)`)

A

0.54

B

0.35

C

0.48

D

0.41

Text Solution

Verified by Experts

The correct Answer is:
C
Decomposition of `NaClO_(3)` is given as: `2NaClO_(3) overset(Delta)rarr underset("(Residue)")(2NaCl) + underset(0.16g)(3O_(2))`
No. of moles of `O_(2)` formed `= (0.16)/(32) = 5 xx 10^(-3), " "n_(NaCl) = (2)/(3) n_(O_(2)) = (2)/(5) xx 5 xx 10^(-3) = (1)/(300)`
`NaCl + Ag^(+) rarr AgCl + Na^(+)` , 1 mole of AgCl is precipitated from one mole of NaCl.
`:.` Mole of `AgCl = (1)/(300) " ":.` Mass of AgCl = Molar mass of `AgCl xx n_(AgCl)`
`= 143.5 xx (1)/(300) square 0.48 g`
Promotional Banner

Similar Questions

Explore conceptually related problems

14.625mg of NaCl is treated with AgNO_(3)(aq) .The mass of AgCl obtained as ppt is ( Ag=108g//mol )

A solution is prepared by adding 10 g of NaCl to 18 g of water. Calculate mole fraction of NaCl. (Molar mass of NaCl = 58.5 g).

If 8g of a non-electrolyte solute is dissolved in 114g of n-octane to educe its vapour pressure to 80% , the molar mass of the solute is [ given molar mass of n-octane is 114g mol^(-1) ]

The specific conductance of saturated solution os silver chloride is k(ohm^(-1)cm^(-1)) . The limiting ionic conductance of Ag^(+) and Cl^(-) ions are x and y respectively. The solubility of AgCl in "gram liter"^(-1) is : (Molar mass of AgCl=143.5"g mol"^(-1) )

5.85 g NaCl is dissolved in 500 ml of water. The molarity is

A 1.0g sample of a pure organic compound cotaining chlorine is fused with Na_(2)O_(2) to convert chlorine to NaCl. The sample is then dissolved in water, and the chloride precipitated with AgNO_(3) , giving 1.96 g of AgCl. If the molecular mass of organic compound is 147, how many chlorine does each molecule contain ?