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3 years ago

What would be the total volume of the new solution when it is changed from 0.2 M to 0.04 M?

2 answers:

34kurt3 years ago

6 0

The question is incomplete.

You need two additional data:

1) the original volume
2) what solution you added to change the volume.

This is a molarity problem, so remember molarity definition and formula:

M = n / V in liters: number of moles per liter of solution

To give you the key to answer this kind of questions, supppose the original volumen was 1 ml and that you added only water (solvent).

The original solution was:

V= 1 ml
M = 0.2 M

Using the formula for molarity, M = n / V

n = M×V = 0.2 M × (1 / 10000)l = 0.0002 moles

For the final solution:

n = 0.0002 moles
M = 0.04

From M = n / V ⇒ V = n / M = 0.002 moles / 0.04 M = 0.05 l

Change to ml ⇒ 0.05 l × 1000 ml / l = 50 ml. This would be the answer for the hypothetical problem that I assumed for you.

I hope this gives you all the cues you need to answer similar problems about molarity.

konstantin123 [22]3 years ago

5 0

Answer:

The answer is 50 mL

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1. If 100 mL of a gas, originally at 760 torr, is compressed to a pressure of 120 kPa

marshall27 [118]

Answer : The final volume will be, 84.4 mL

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 760 torr

$P_2$ = final pressure = 120 kPa = 900 torr (1 kPa = 7.5 torr)

$V_1$ = initial volume = 100 mL

$V_2$ = final volume = ?

Now put all the given values in the above equation, we get:

$760torr\times 100mL=900torr\times V_2$

$V_2=84.4mL$

Therefore, the final volume will be, 84.4 mL

3 0

3 years ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

Answer: The mass of glucose in final solution is 0.420 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$