All categories

2 years ago

Determine the percent composition by mass of a 100g salt solution which contains 20g salt

Chemistry

2 answers:

prisoha [69]2 years ago

8 0

Answer:

Mass of solution=100g

mass of salt=20g

so; mass of solute=80g

percentage composition =(mass of salt/total

mass) ×100

= \frac{20}{100} \times 100 \\ = 20\%

glad to help you

hope it helps

lara [203]2 years ago

8 0

The percent composition by mass of a 100g salt solution which contains 20g salt is $\bold{16.6\%}$

<u>Solution:</u>

Percent composition is calculated as, weight of the solute divided by amount of solute added to 100 g is solution multiplied by 100.

$\Rightarrow \frac{\text {weight of solute}(g)}{\text {weight of solute}(g)+\text {weight of solution(g)}} \times 100$

Given, 100 g of salt solution with 20 g of salt

Weight of the solute and solvent is $20 + 100 g = 120 g$

$percent composition=\frac{20}{120} \times 100=\frac{200}{12}=16.6$

$\therefore \text { percent composition } =16.6 \%$

You might be interested in

How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C

NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0

2 years ago

The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol. What mass of O2, in grams, is required t

Ira Lisetskai [31]

The reaction formula of this is C3H8 + 5O2 --> 3CO2 + 4H2O. The ratio of mole number of C3H8 and O2 is 1:5. 0.025g equals to 0.025/44.1=0.00057 mole. So the mass of O2 is 0.00057*5*32=0.0912 g.

6 0

2 years ago

How much water(in grams) at its boiling point can be vaporized by adding 1.50 kJ of heat? The molar heat of vaporization for wat

LekaFEV [45]

Answer:

0.66g of water

Explanation:

Molar heat of vaporization of any substance is defined as the heat necessary to vaporize 1 mole of the substance.

If heat of vaporization of water is 40.79kJ/mol and you add 1.50kJ, the moles you vaporize are:

1.50kJ × (1mol / 40.79kJ) = 0.0368 moles of water.

As molar mass of water is 18.01g/mol, mass of water that can be vaporized are:

0.0368 moles × (18.01g / mol) = <em>0.66g of water</em>

6 0

2 years ago

A sample of fluorine gas is confined in a 5.0-L container at 0.432 atm and 37 °C. How many moles of gas are in this sample?

Ahat [919]

Answer:

About 0.08486 moles

Explanation:

PV=nRT, when P is the pressure in atmospheres, V is the volume in liters, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin.

$0.432 \cdot 5= n \cdot 0.0821 \cdot 310$

$n\approx 0.08486$ moles

Hope this helps!

7 0

2 years ago

The concentration of Si in an Fe-Si alloy is 0.25 wt%. What is the concentration in kilograms of Si per cubic meter of alloy?

kodGreya [7K]

Answer : The concentration of Si in kilograms is, $19.55kg/m^3$

Explanation :

As we are given that, the concentration of Si in an Fe-Si alloy is 0.25 wt% that means:

Weight of Si = 0.25 g = 0.00025 kg

Weight of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg

Density of Si = $2.32g/cm^3=2.32\times 10^6g/m^3$

Density of Fe = $7.87g/cm^3=7.87\times 10^6g/m^3$

Now we have to calculate the concentration in kilograms of Si per cubic meter of alloy.

Concentration of Si in kilograms = $\frac{\text{Weight of Si in 100 g of alloy}}{\text{Volume of 100 g of alloy}}$

Concentration of Si in kilograms = $\frac{\text{Weight of Si in 100 g of alloy}}{\frac{\text{Wight of Fe}}{\text{Density of Fe}}+\frac{\text{Wight of Si}}{\text{Density of Si}}}$

Now put all the given values in this expression, we get:

Concentration of Si in kilograms = $\frac{0.00025kg}{\frac{99.75g}{7.87\times 10^6g/m^3}+\frac{0.25g}{2.23\times 10^6g/m^3}}$

Concentration of Si in kilograms = $19.55kg/m^3$

Thus, the concentration of Si in kilograms is, $19.55kg/m^3$

5 0

2 years ago