Question

If 15.6 g of hydrate are heated and only 11.7 g of anhydrous salt remain, calculate the % of water lost.

Answer 1

Answer:

Percent loss of water = 25%

Explanation:

Given data:

Mass of hydrated salt = 15.6 g

Mass of anhydrous salt = 11.7 g

Percentage of water lost = ?

Solution:

First of all we will calculate the mass of water in hydrated salt.

Mass of water =  Mass of hydrated salt - Mass of anhydrous salt

Mass of water = 15.6 g - 11.7 g

Mass of water = 3.9 g

Now we will calculate the percentage.

Percent loss of water = mass of water / total mass × 100

Percent loss of water = 3.9 g/ 15.6 g × 100

Percent loss of water = 25%

Answer 2

The percentage of water lost will be "25%".

According to the question,

Mass,

• Hydrate salt = 15.6 g
• Anhydrous salt = 11.7 g

The mass of water will be:

= $Mass \ of \ hydrate \ salt - Mass \ of \ anhydrous \ salt$

= $15.6-11.7$

= $3.9 \ g$

hence,

The water lost will be:

= $\frac{Mass \ of \ water}{Total \ mass}\times 100$

By substituting the values, we get

= $\frac{3.9}{15.6}\times 100$

= $25$ (%)

Thus the above solution is correct.

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