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FrozenT [24]
3 years ago
6

If the standard solutions had unknowingly been made up to be 0.0024 m agno3 and 0.0040 m k2cro4, would this have affected your r

esults? how?
Chemistry
2 answers:
Mnenie [13.5K]3 years ago
6 0
2AgNO3+K2CrO4⇒Ag2CrO4(s)+2KNO3
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with 
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04
OlgaM077 [116]3 years ago
6 0

If the standard solutions had unknowingly been made up to be 0.0024 M AgNO₃ and 0.0040 M K₂CrO₄ the chemical equation would be:  

2AgNO₃ + K₂CrO₄ → Ag₂CrO₄ (s) + 2KNO₃

By mixing 0.0024 M AgNO₃ and 0.004 M K₂CrO₄ one would have Ag₂CrO₄ precipitated out.  

Thus, one is left with 0.0024 M KNO₃ mixed with (0.004 -0.0024 / 2) M = 0.0028 M of K₂CrO₄.  

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H=1.01
meriva

Answer:

Molality of solution=10.11 m

Explanation:

We are given that

Given mass of KCl(WB)=75.3 g

Given mass of water (WA)=100 g=100/1000=0.1 kg

1 kg=1000 g

Molar mass of H=1.01 g

Molar mass of K=39g

Molar mass of Cl=35.45 g

We have to find the molality of a solution.

Molar mass of KCl(MB)=39+35.45

Molar mass of KCl(MB) =74.45 g

Number of moles of solute (KCl)=\frac{given\;mass}{molar\;mass}=\frac{W_B}{M_B}

Number of moles of solute (KCl)=\frac{75.3}{74.45}

Number of moles of solute (KCl)=1.011 moles

Molality of solution

=\frac{number\;of\;moles\;of\;solute}{mass\;of\;solvent}

Using the formula

Molality of solution=\frac{1.011}{0.1}

Molality of solution=10.11 m

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