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ArbitrLikvidat [17]
3 years ago
11

A sample of hydrate sodium carbonate has a mass of 71.5 g. after heating, the anhydrous sodium carbonate has a mass of 26.5 gram

s. Calculate the % water.
Chemistry
1 answer:
drek231 [11]3 years ago
3 0

Answer:

62.9%

Explanation:

Step 1: Given data

  • Mass of hydrate sodium carbonate (mNa₂CO₃.xH₂O): 71.5 g
  • Mass of anhydrous sodium carbonate (mNa₂CO₃): 26.5 g

Step 2: Calculate the mass lost of water

We will use the following expression.

mH₂O = mNa₂CO₃.xH₂O - mNa₂CO₃

mH₂O = 71.5 g - 26.5 g = 45.0 g

Step 3: Calculate the percent of water in the hydrate sodium carbonate

We will use the following expression.

%H₂O = mH₂O / mNa₂CO₃.xH₂O × 100%

%H₂O = 45.0 g / 71.5 g × 100%

%H₂O = 62.9%

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1. 100gm of a 55% (M/M) nitric acid solution is to be diluted to 20% (M/M) nitric acid.
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Answer:

The volume of water to be added is 0.175 liters of water

Explanation:

The given concentration of the nitric acid = 55% (M/M)

The mass of the nitric acid solution = 100 gm

The concentration solution is to diluted to = 20% (M/M)

The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution

Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get

Let "x" represent the volume of the resulting solution, we have;

20% of x = 55 g of nitric acid

∴ 20/100 × x = 55 g

x = 55 g × 100/20 =  275 g

The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid

The mass of extra water to be added = 275 g - 100 g = 175 g

Volume = Mass/Density

The density of water ≈ 1 g/ml

∴ The volume of water to be added that gives 175 g of water =  175 g/(1 g/ml) = 175 ml. = 0.175 l

The volume of water to be added = 0.175 liters of water.

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