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Yuki888 [10]
3 years ago
10

What is the molarity of a solution made with 64 grams of sodium hydroxide in 4 liters of water

Chemistry
1 answer:
ehidna [41]3 years ago
7 0

Answer:

The molarity of the solution: 0,27M

Explanation:

First , we calculate the weight of 1 mol of NaCl:

Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol

58,5 g---1 mol NaCl

64 g--------x= (64 g x1 mol NaCl)/58,5 g= 1, 09 mol NaCl

A solution molar--> moles of solute in 1 L of solution:

4 L-----1,09 mol NaCl

1L----x0( 1L x1,09 mol NaCl)/4L =0,27moles NaCl--->0,27M

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An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?
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<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

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The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

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

where,

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M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

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