0.1 M preparation of NaCl and NaOH in 100 ml water.

1.	0.1 M preparation of NaCl and NaOH in 100 ml water.
Answer» ANSWER: We KNOWN that normality is the gram equivalent WEIGHT of solute per unit volume . Also gram equivalent weight (x) is calculated as follow . X= molecular weight of NAOH / acidity of NAOH . 1000ml -----1N ----- 40 G . 100 ml ---- 0.1 ---X . From above we can see that to prepare 1000ml , 1N NAOH solution require 40g of NAOH . So to prepare 100 ml of 0.1 N NAOH solution require Xg of NAOH calculated as follow. From l 1000ml × 1N × Xg = 40 g × 100ml × 0.1N . X = 40g × 100ml × 0.1 × 1000ml × 1N . X = 0.4 g NAOH.

Answer»

We KNOWN that normality is the gram equivalent WEIGHT of solute per unit volume .

Also gram equivalent weight (x) is calculated as follow .

X= molecular weight of NAOH / acidity of NAOH .

1000ml -----1N ----- 40 G .

100 ml ---- 0.1 ---X .

From above we can see that to prepare 1000ml , 1N NAOH solution require 40g of NAOH .

So to prepare 100 ml of 0.1 N NAOH solution require Xg of NAOH calculated as follow.

From l

1000ml × 1N × Xg = 40 g × 100ml × 0.1N .

X = 40g × 100ml × 0.1 × 1000ml × 1N .

X = 0.4 g NAOH.

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