Question

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How many liters of water would be needed to dissolve 21.6 g of lithium nitrate to make a 1.3 M (molar) solution?

What is the molarity of a solution made of 215.1 g of HCl is dissolved to make 2.0 L of solution?

How much concentrated 18 M H
`x_(2)`SO
`x_(4)` is needed to prepare 250.0 mL of a 6.0 M solution?

A chemist has a stock solution of HBr that is 10.0 M and would like to make 450.0 mL of 3.0 M HBr, how would he/she do it?

How much water should be added to 50.0 mL of 12 M hydrochloric acid to make a 4.0 M solution?

Lulu Labwrecker carefully pipeta 25.0 mL of 0.525 M NaOH into a test tube. She places the test tube into a small beaker to keep it from spilling and then pipets 75.0 mL of 0.355 M HCl into another test tube. When Lulu reaches out to put this test tube of acid into the beaker along with test tube of base she accidentally knocks the test tubes together hard enough to break them and their respective contents combine in the bottom of the beaker. Is the solution formed from the contents of the two test tubes acidic or basic? What is the pH of the resulting solution?

Answer 1

1. 0.240 liters of water would be needed to dissolve 21.6 g of lithium nitrate to make a 1.3 M (molar) solution.

2. 2.9 M is the molarity of a solution made of 215.1 g of HCl is dissolved to make 2.0 L of solution.

3.83.3 ml of concentrated 18 M H2SO4 is needed to prepare 250.0 mL of a 6.0 M solution.

4. 135 ml of stock HBr will be required to dilute the solution.

5. 150 ml of water should be added to 50.0 mL of 12 M hydrochloric acid to make a 4.0 M solution

6. The pH of the resulting solution is 13.89

Step-by-step explanation:

The formula used in solving the problems is

number of moles=
`(mass)/(atomic mass of one mole)` 1st equation

molarity =
`(number of moles)/(volume)` 2nd equation

Dilution formula

M1V1 = M2V2 3rd equation

1. Data given

mass of Lithium nitrate = 21.6 grams

atomic mass of on emole lithium nitrate = 68.946 gram/mole

Molarity is given as 1.3 M

VOLUME=?

Calculate the number of moles using equation 1

n =
`(21.6)/(68.946)`

= 0.313 moles of lithium nitrate.

volume is calculated by applying equation 2.

volume =
`(0.313)/(1.3)`

= 0.240 litres of water will be used.

2. Data given:

mass of HCl = 215.1 gram

atomic mass of HCl = 36.46 gram/mole

volume = 2 litres

molarity = ?

using equation 1 number of moles calculated

number of moles =
`(215.1)/(36.46)`

number of moles of HCl = 5.899 moles

molarity is calculated by using equation 2

M =
`(5.899)/(2)`

= 2.9 M is the molarity of the solution of 2 litre HCl.

3. data given:

molarity of H2SO4 = 18 M

Solution to be made 250 ml of 6 M

USING EQUATION 3

18 x V1= 250 x 6

V1 = 83.3 ml of concentrated 18 M H2SO4 will be required.

4. data given:

M1= 10M, V1 =?, M2= 3 ,V2= 450 ml

applying the equation 3

10 x VI = 3x 450

V1 = 135 ml of stock HBr will be required.

5. Data given:

V1 = 50 ml

M1= 12 M

V2=?

M2= 4

applying the equation 3

50 x 12 = 4 x v2

V2 = 150 ml.

6. data given:

HCl + NaOH ⇒ NaCl + H20

molarity of NaOH = 0.525 M

volume of NaOH = 25 ml

molarity of acid HCl= 75 ml

volume of HCl = 0.335 ml

pH=?

Number of moles of NaOH and HCl is calculated by using equation 1 and converting volume in litres

moles of NaOH = 0.0131

moles of HCl= 0.025 moles

The ratio of moles is 1:1 . To find the unreacted moles of acid and base which does not participated in neutralization so the difference of number of moles of acid minus number of moles of base is taken.

difference of moles = 0.0119 moles ( NaOH moles is more)

Molarity can be calculated by using equation 1 in (25 +75 ml) litre of solution

molarity =
`(0.0119)/(0.1)`

= 0.11 M (pOH Concentration)

14 = pH + pOH

pH = 14 - 0.11

pH = 13.89