All categories

3 years ago

Calculate the molarity when 135 g of KCl is dissolved to make a 2.50 L solution. Round to two significant digits.

Chemistry

1 answer:

inysia [295]3 years ago

7 0

Explanation:

$\textrm{Molarity of a solution}=\frac{\textrm{Number of moles of solvent}}{\textrm{Volume of solution in Liters}}$

We are given a $2.50L$ solution which contains $135g$ of $KCl$ .

To calculate number of moles of $KCl$ present, we need to find it's molar mass from charts. Molar Mass of $KCl$ is known to be $74.55\frac{g}{Mol}$ .

Number of moles of $KCl$ present = $\frac{\textrm{Given weight}}{\textrm{Molar Mass}}\textrm{ = }\frac{135g}{74.55\frac{g}{Mol}}\textrm{ = }1.81087mol$

Molarity = $\frac{1.81087mol}{2.50L}=0.7243\frac{mol}{L}$

∴ Molarity of given $KCl$ solution = $0.72\frac{mol}{L}$

You might be interested in

For a first-order reaction, A → B, the rate coefficient was found to be 3.4 × 10-4 s-1 at 23 °C. After 5.0 h, the concentration

Illusion [34]

Answer:

the original concentration of A = 0.0817092 M

Explanation:

A reaction is considered to be of first order it it strictly obeys the graphical equation method.

$k_1 = \dfrac{2.303}{t}log \dfrac{a}{a-x}$

where;

k = the specific rate coefficient = 3.4 × 10⁻⁴ s⁻¹

t = time = 5.0 h = 5.0 × 3600 = 18000 seconds

a = initial concentration = ???

a - x = remaining concentration of initial concentration at time t = 0.00018 mol L⁻¹

$3.4 \times 10^{-4}= \dfrac{2.303}{18000}log \dfrac{a}{0.00018}$

$3.4 \times 10^{-4}= 1.27944 \times 10^{-4} \times log \dfrac{a}{0.00018}$

$\dfrac{3.4 \times 10^{-4}}{1.27944 \times 10^{-4}}= log \dfrac{a}{0.00018}$

$2.657= log \dfrac{a}{0.00018}$

$10^{2.657}= \dfrac{a}{0.00018}$

$453.94 = \dfrac{a}{0.00018}$

$a =453.94 \times 0.00018$

a = 0.0817092 M

Thus , the original concentration of A = 0.0817092 M

8 0

3 years ago

The concentration of hydrogen peroxide solution can be determined by

max2010maxim [7]

The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

with acidified potassium manganate(VII) solution. In this reaction the hydrogen

peroxide is oxidised to oxygen gas.

A 5.00 cm3 sample of the hydrogen peroxide solution was added to a volumetric flask

and made up to 250 cm3 of aqueous solution. A 25.0 cm3 sample of this diluted

solution was acidified and reacted completely with 24.35 cm3 of 0.0187 mol dm–3

potassium manganate(VII) solution.

Write an equation for the reaction between acidified potassium manganate(VII)

solution and hydrogen peroxide.

Use this equation and the results given to calculate a value for the concentration,

in mol dm–3, of the original hydrogen peroxide solution.

(If you have been unable to write an equation for this reaction you may assume that

3 mol of KMnO4 react with 7 mol of H2O2. This is not the correct reacting ratio.)

Answer:

2.275 M

Explanation:

The equation of the reaction is;

2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)

Let;

CA= concentration of MnO4^- = 0.0187 mol dm–3

CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution = 5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

V2 = final volume of H2O2 solution = 250 cm3

C1 = C2V2/V1

C1 = 0.0455 * 250/5

C1 = 2.275 M

8 0

3 years ago

A gas has a volume of 25.0 mL when under a pressure of 525 mmHg. What is the new pressure when the volume has been increased to

Fudgin [204]

Answer:

152.26 mmHg

Explanation:

pv=p'v'

525× 25=p'×86.2

p'=525×25÷ 86.2

p'=152.26

5 0

3 years ago

An airplane travels 2100 km at 1000km/hE. It encounters a wind and slows to 800 km/h E for the next 1300 km. What is the average

Deffense [45]

Answer:

The average velocity of the airplane for this trip is 1684.21 km/h

Explanation:

Average velocity is the rate of change of displacement with time. That is,

Average velocity = $\frac{Displacement }{Change in time}$ = Δx / Δt = $\frac{x2 - x1}{t2 - t1}$

Now we will calculate the time taken by the airplane for the first motion before it encounters a wind.

From,

Velocity = $\frac{Distance traveled}{Time taken}$

Time = $\frac{Distance traveled}{Velocity}$

Therefore, Time = $\frac{2100km }{1000km/h}$

Time = 2.1h

This is the time taken before the airplane encounters a wind.

Hence, t1 = 2.1h

Now, For the time taken by the airplane when it encounters a wind

Also from,

Velocity = $\frac{Distance traveled}{Time taken}$

Time = $\frac{Distance traveled}{Velocity}$

Therefore, Time = $\frac{1300km }{800km/h}$

Time = 1.625h

Hence, t2 = 1.625h

Now, to calculate the average velocity

Average velocity = $\frac{x2 - x1}{t2 - t1}$

x1= 2100, x2= 1300, t1= 2.1h and t2= 1.625h

Hence, Average velocity = $\frac{1300 - 2100}{1.625 - 2.1}$

Average velocity = 1684.21 km/h

7 0

3 years ago

What is chemical change? what are some ways to tell chemical change happened?

andrew-mc [135]

Answer:

A chemical change occurs when the reactants chemical compositions have changed

Explanation:

a change in colour, change in temperature, change in smell, formation of a precipitate, or the formation of gas bubbles

Hopes this helps

4 0

2 years ago