Calculate the molarity of a solution prepared by dissolving 11.5 g of solid N_2O

Question

Calculate the molarity of a solution prepared by dissolving 11.5 g of solid N_2OH is enough water to make 1.50 L of solution. Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl is enough water to make 26.8 mL of solution Give the concentration of each type of ion in the following solution 0.50 M Co(NO)_2 I M Fe(ClO_4)_3 What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H_2SO_4 solution? Typical blood serum is about 0.14 M NaCl. What volume of blood contains 1.0 mg NaCl? 47.60 mL of a 1.80 M solution is diluted to a volume of 278 mL. A 139 ml portion of that solution is diluted using 139 mL of water. What is the final concentration? Given a 0.75 M solution of AgNO_3(aq), what is the mass of Ag+ in a 175 ml sample of that solution? MW of Ag = 107.86 g/mole, MW of N = 14 g/mole, MW of O = 15.99 g/mole

Explanation / Answer

4.1 Molarity = Moles of solute / volume of solution in litres

Moles of solute = mass of solute dissolved / Mol wt of solute

Mass of solute = 11.5 g

Mol wt of solute NaOH = 40 g

Moles = 11.5 / 40 = 0.288

Molarity = moles / volume of solution = 0.288 / 1.5 = 0.192 M

2) Molarity = moles / Volume in litres

Volume = 26.8 mL = 26.8 / 1000 L

Moles = MAss / mol wt = 1.56 / 36.5 = 0.0427

Molarity = 0.0427 X 1000 / 26.8 = 1.59 M

3)

a) Co(NO3)2 Each molecule will give one ion of Co+2 and two ions of NO3-

[CO+2] = 0.5 M

[NO3- ] = 2 X 0.5 = 1 M

b) Fe(ClO4)3

Each molecule will give one ion of Fe+3 and three ions of ClO4-

[Fe+3] = 1 M

[ClO4-] = 3X1 = 3M

4) we will use the formula

M!V1 = M2V2

M1= 16M

V1 = ? volume needed

M2 =0.1 M

V2 = 1.5 L

V1= M2V2 / M1 = 0.1 X 1.5 / 16 = 0.00938 Litres = 9.38 mL

5) molarity = 0.14 M = 0.14 moles / Litres

1 mole = 23 + 35.5 = 58.5 g

So 0.14 moles = 58.5 X 0.14 grams = 8.19 grams / Litre

So for 1 gram we need = 1/8.19 Litres of solution = 0.122 Litres = 122mL

so for 1mg we need = 122 / 1000 mL = 0.122 mL

47)

V1 = 60mL

M1 = 1.8 M

V2 = 278 mL

M2 = ?

M1V1 = M2V2

M2 = M1V1 / V2 = 60 x 1.8 / 278 = 0.388 Molar

After further dilution

M3 = ?

V3 = 139 + 139 = 278mL

M2 = 0.388

V2 = 139mL

V3 = M2V2 / M3 = 0.388 X 278 / 139 = 0.776 molar

Final concentration

48)

the concentration of AgNO3 = 0.75 M

so 0.75 moles of AgNO3 in 1 litre

So moles of Ag+ in 1Litres = 0.75 mole

Moles in 1mL = 0.75 / 1000

Moles in 170mL = 0.75 X 170 /1000= 0.113 moles

atomic weight of Ag= 107.86 g / mole

so mass of Ag in 0.113 moles = 107.86 X 0.113 = 12.188 grams

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