5. Calculate the molar concentration of solute in each of the following solution

Question

5. Calculate the molar concentration of solute in each of the following solutions: (a) Exactly 7.59 g of NH3 dissolved in 1.500 L of solution. (b) Exactly 0.291 g of H2S04 dissolved in 8.65 L of solution. (c) Exactly 85.2 g of NaOH dissolved in 5.00 L of solution. (d) Exactly 31.25 g of NaCl dissolved in 2.00 L of solution. (e) Exactly 6.00 mg of atmospheric N2 dissolved in water in a total solution volume of 2.00 L () Exactly 1,000 g of ethylene glycol, C-HoO2 dissolved in water and made up to a volume of 2.00 L . mass rsinle molaluhy s 6. If one wished to prepare 2 liters of 1.2 M NaCI, how many liters of 2M NaCl would be required? 7. How much of a standard solution of a sodium compound containing 1000 mg of Na* per liter should be taken to prepare 5 liters of a standard sodium solution containing 10 mg of sodium per liter? 8. Exactly 1.50 liters of 2.00 M HCI was added to a 2.00-liter volumetric flask, and the volume brought to 2.00 liters with water. What was the molar concentration of HCl in the resulting solution? 9. Exactly 1.00 liter of 1.00M HCI was added to a 2.00-liter volumetric flask. Next, a total of 0.25 mole of solid NaOH was added. After the chemical reaction between the HCI and the NaOH was complete, water was added to bring the

Explanation / Answer

5)

a)

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass(NH3)= 7.59 g

use:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(7.59 g)/(17.03 g/mol)

= 0.4456 mol

volume , V = 1.5 L

use:

Molarity,

M = number of mol / volume in L

= 0.4456/1.5

= 0.2971 M

Answer: 0.297 M

b)

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass(H2SO4)= 0.291 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(0.291 g)/(98.09 g/mol)

= 2.967*10^-3 mol

volume , V = 8.65 L

use:

Molarity,

M = number of mol / volume in L

= 2.967*10^-3/8.65

= 3.43*10^-4 M

Answer: 3.43*10^-4 M

c)

Molar mass of NaOH,

MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

mass(NaOH)= 85.2 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(85.2 g)/(40 g/mol)

= 2.13 mol

volume , V = 5 L

use:

Molarity,

M = number of mol / volume in L

= 2.13/5

= 0.426 M

Answer: 0.426 M

d)

Molar mass of NaCl,

MM = 1*MM(Na) + 1*MM(Cl)

= 1*22.99 + 1*35.45

= 58.44 g/mol

mass(NaCl)= 31.25 g

use:

number of mol of NaCl,

n = mass of NaCl/molar mass of NaCl

=(31.25 g)/(58.44 g/mol)

= 0.5347 mol

volume , V = 2 L

use:

Molarity,

M = number of mol / volume in L

= 0.5347/2

= 0.2674 M

Answer: 0.267 M

Only 4 parts at a time

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