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Scilla [17]
3 years ago
10

What is the molarity of a solution that contains 14.92 grams magnesium oxide (MgO) in 365ml of solution?

Chemistry
2 answers:
AnnZ [28]3 years ago
6 0

Answer:

Option A. 1.01 molar solution

Explanation:

Step 1:

Data obtained from the question. This include the following:

Mass of MgO = 14.92g

Volume of solution = 365mL

Molarity of solution =.?

Step 2:

Determination of the number of mole present in 14.92g of MgO.

This is illustrated below:

Mass of MgO = 14.92g

Molar mass of MgO = 24.3 + 16 = 40.3g/mol

Number of mole MgO =...?

Mole = Mass /Molar Mass

Number of mole of MgO = 14.92/40.3 = 0.370 mole

Step 3:

Determination of the molarity of the solution.

This can be obtained as follow:

Mole of MgO = 0.370 mole

Volume = 365mL = 365/1000 = 0.365L

Molarity =..?

Molarity = mole /Volume

Molarity = 0.370/0.365

Molarity = 1.01M

Yuki888 [10]3 years ago
3 0

Answer:

a. 1.01 molar solution

Explanation:

Now we have to recall the formula;

n= CV

Where

n= number of moles of magnesium oxide

C= concentration of magnesium oxide solution = ???

V= volume of magnesium oxide solution= 365 ml

But;

n= m/M

Where;

m= given mass of magnesium oxide= 14.2 g

M= molar mass of magnesium oxide= 40.3044 g/mol

Substituting values;

14.2g/40.3044 g/mol = C×365/1000

0.3523= 0.365 C

C= 0.3523/0.365

C= 0.965

C= 1.00 molar (approx)

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A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som
irinina [24]

Answer:

m_{KBr}=6.030gKBr

Explanation:

Hello.

In this case, since the normal boiling point of X is 117.80 °C, the boiling point elevation constant is 1.48 °C*kg*mol⁻¹, the mass of X is 100 g and the boiling point of the mixture of X and KBr boils at 119.3 °C, we can use the following formula:

(T_b-T_b_0)=i*m*K_b

Whereas the Van't Hoff factor of KBr is 2 as it dissociates into potassium cations and bromide ions; it means that we can compute the molality of the solution:

m=\frac{T_b-T_b_0}{i*K_b}=\frac{(119.3-117.8)\°C}{2*1.48\°C*kg*mol^{-1}}\\  \\m=0.507mol/kg

Next, given the mass of solventin kg (0.1 kg from 100 g), we compute the moles KBr:

n_{KBr}=0.507mol/kg*0.1kg=0.0507mol

Finally, considering the molar mass of KBr (119 g/mol) we compute the mass that was dissolved:

m_{KBr}=0.0507mol*\frac{119g}{1mol} \\\\m_{KBr}=6.030gKBr

Best regards.

3 0
3 years ago
List the following aqueous solutions in order of decreasing freezing point: 0.040 m glycerin (C3H8O3), 0.020 m potassium bromide
gogolik [260]

Answer:

Solutions from highest to lowest freezing point:

0.040 m glycerin = 0.020 m potassium bromide > 0.030 m phenol

Explanation:

\Delta T_f=i\times K_f\times m

\Delta T_f = Depression in freezing point

i = van'T Hoff fcator

K_f = Molal depression constant of solvent

m = molality of the solution

Higher the value of depression in freezing point at  lower will be freezing temperature the solution.

1. 0.040 m glycerin

Molal depression constant of water = K_f=1.86^oC/m

i = 1 ( organic molecule)

m = 0.040 m

\Delta T_{f,1}=1\times\times 1.86^oC/m\times 0.040 m

\Delta T_{f,1}=0.0744^oC

2. 0.020 m potassium bromide

Molal depression constant of water = K_f=1.86^oC/m

i = 2 (ionic)

m = 0.020 m

\Delta T_{f,2}=2\times\times 1.86^oC/m\times 0.020 m

\Delta T_{f,2}=0.0744^oC

3. 0.030 m phenol

Molal depression constant of water = K_f=1.86^oC/m

i = 1 (organic)

m = 0.030 m

\Delta T_{f,3}=1\times\times 1.86^oC/m\times 0.030 m

\Delta T_{f,3}=0.0558^oC

0.0744^oC=0.0744^oC > 0.0558^oC

\Delta T_{f,1}=\Delta T_{f,2}>\Delta T_{f,3}

Solutions from highest to lowest freezing point:

0.040 m glycerin = 0.020 m potassium bromide > 0.030 m phenol

5 0
3 years ago
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