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ira [324]
3 years ago
12

Determine the molarity for each of the following solution solutions:

Chemistry
1 answer:
____ [38]3 years ago
5 0

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of H_2SO_4 solution is, 0.00525 mole/L

(c)The molarity of Al(NO_3)_3 solution is, 0.0612 mole/L

(d)The molarity of CuSO_4.5H_2O solution is, 7.61 mole/L

(e)The molarity of Br_2 solution is, 0.0565 mole/L

(f)The molarity of C_2H_5NO_2 solution is, 0.0113 mole/L

Explanation :

<u>(a) 1.457 mol of KCl in 1.500 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Solute is KCl.

\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L

The molarity of KCl solution is, 0.9713 mole/L

<u>(b) 0.515 gram of H_2SO_4, in 1.00 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Solute is H_2SO_4

Molar mass of H_2SO_4 = 98 g/mole

\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L

The molarity of H_2SO_4 solution is, 0.00525 mole/L

<u>(c) 20.54 g of Al(NO_3)_3 in 1575 mL of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Solute is Al(NO_3)_3

Molar mass of Al(NO_3)_3 = 213 g/mole

\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L

The molarity of Al(NO_3)_3 solution is, 0.0612 mole/L

<u>(d) 2.76 kg of CuSO_4.5H_2O in 1.45 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Solute is CuSO_4.5H_2O

Molar mass of CuSO_4.5H_2O = 250 g/mole

\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L

The molarity of CuSO_4.5H_2O solution is, 7.61 mole/L

<u>(e) 0.005653 mol of Br_2 in 10.00 ml of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

Solute is Br_2.

\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L

The molarity of Br_2 solution is, 0.0565 mole/L

<u>(f) 0.000889 g of glycine, C_2H_5NO_2, in 1.05 mL of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Solute is C_2H_5NO_2

Molar mass of C_2H_5NO_2 = 75 g/mole

\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L

The molarity of C_2H_5NO_2 solution is, 0.0113 mole/L

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The original temperature of the gas, T₁ = 263.0 K

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The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

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