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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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In order to do this, we need various data. First to all, we need tje molecular mass of the ethanol. this can be obtained in handbooks, or simply taking the atomic weights of carbon (12 g/mol), Hydrogen (1 g/mol) and oxygen (16 g/mol) and summing those values:

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<h2>%m/m = 0.9975%</h2>

To get the mole fraction, we first need to get the volume of solvent. From the density, we can get the mass of solution:

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