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Direct Combination

0.001 mole of...

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0.001 mole of Cr(NH3)5(NO3)SO4 was passed through cation exchanger and the acid coming out of it required 20 ml 0.1M NaOH for neutralisation. Hence the complex is ?

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Q. 0.001 mol of

C o (N H_{3})_{5} (N O_{3}) (S O_{4})

was passed through a cation exchanger and the acid coming out of its required 20 ml of 0.1 M NaOH for neutralisation. Hence, the complex is :

0.001

[C o (N H_{3})_{5} (N O_{3}) (S O_{4})]

was passed through a cation exchanger and the acid coming out of it required

20

0.1 M N a O H

for neutralisation. Thus, the complex is :

Q. 0.0012 mole of

C r C l_{3} \cdot 6 H_{2} O

was passed through a cation exchange resin and acid coming out of it required 28.5 mL of 0.125 M

N a O H

. Hence, the formula of the complex is:

Q. A solution containing

0.319 g

C r C l_{3} .6 H_{2} O

was passed through cation exchanger and the solution given out was neutralised by

28.5 m L

0.125 M N a O H

. What is the correct formula of complex ?

Q. 100 ml of sample of hard water is passed through a column of the cation exchange resin

H_{2} R

. The water coming out of the column requires 20 ml of 0.025 M

N a O H

for the titration. The hardness of water expressed as ppm of

C a^{2 +}

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