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mezya [45]
3 years ago
11

What is the percentage of water in hydrated calcium chloride​

Chemistry
2 answers:
earnstyle [38]3 years ago
8 0

Answer:

24.5%

Explanation:

You just add up the atomic masses.  

Ca - 40.078  

Cl2 - 35.4527 x 2 = 70.9054  

------ 110.9834  

H4 - 1.00794 x 4 = 4.03176  

O2 - 31.9998  

------ 36.03056  

TOTAL - 147.01396  

So the water is 36.03056/147.01396 = .245082576 but that is only accurate to three decimals (because the mass of Ca was only given to three decimals) so we write .245 and that is 24.5%

This is not my answer but I found it on Yahoo answers and it was answered by Anonymous.

oee [108]3 years ago
8 0

Answer: <u>24.5%</u>

Explanation: Molar mass of CaCl2 * 2 H2O = 40.078 + ( 2 x 35.453) + ( 2 x 18.02)=147.024 g/mol  

% H2O = 2 x 18.02 x 100/ 147.024 =24.5

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The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be
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<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

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Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC

Hence, the freezing point of solution is 5.35°C

3 0
3 years ago
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