If laundry bleach is an equimolar mixture of NaCl and NaClO and a 10.0-mL sample

Question

If laundry bleach is an equimolar mixture of NaCl and NaClO and a 10.0-mL sample (density = 1.01 g/mL) of bleach solution requires 41.0 mL of 0.105 M AgNO3 to precipitate all of the chloride ion, calculate the % (w/v) of NaClO in the solution. Also can you explain what w/v is? If laundry bleach is an equimolar mixture of NaCl and NaClO and a 10.0-mL sample (density = 1.01 g/mL) of bleach solution requires 41.0 mL of 0.105 M AgNO3 to precipitate all of the chloride ion, calculate the % (w/v) of NaClO in the solution. Also can you explain what w/v is?

Explanation / Answer

Given

Molar mass of NaClO = 74.44 g/mol

Volume = = 41.0 mL = 0.041 L

Molarity (M) = 0.105 M

Let’s write the balance reaction of NaCl & NaClO with AgNO3

NaCl + AgNO3 --> AgCl + NaNO3

2NaClO + 2AgNO3 --> 2AgCl + 2NaNO3 + O2

In both the above case NaCl and NaClO have 1 mole of Cl per mole

Let’s calculate moles of Cl-

Here

Molarity = 0.105 M AgCl

Volume = 41 mL = 0.041 L

# mole = Molarity * Volume = 0.105M * 0.041 L = 0.004305 moles of Cl-

Here 0.004305/2 moles of each of two NaCl & NaClO in 10.0 mL

= 0.00215 moles in 10.0 mL

Therefore in

in 100 mL 0.021525 moles

(1000 mL) 1.0 Lsample contain

# moles = 0.2152 moles of NaClO

We know

# moles = Mass/ Molar mass

Mass = # moles * Molar mass = 0.02152 moles * 74.44 g/mol

= 1.6019 g / 100 mL

Answer % (w/v) of NaClO in the solution = 1.6019 %( W/V)

Here

W = weight in gram (g)

V = Volume in mL

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