4. The molar heat of vaporization Molar mass/g mol 159.81 of bromine at 25°C is

Question

4. The molar heat of vaporization Molar mass/g mol 159.81 of bromine at 25°C is +15.0 kJ-mol-I. How much heat (in kJ) would be required to vaporize 25.0 g of bromine, Br? (A) 375 k (C) 2.35 kJ Br2 (B) 95.9 kJ (D) 1.67 kJ 5. What is the molar concentration Molar mass/gmo Na-CO. 0599 of a 12.0% by mass sodium Na carbonate, Na COs, aqueous solution having a density of 1.124 g mL-1? (A) 1.44 M (C) 1.13 M (B) 1.27 M (D) 1.01 M 6. A solution is prepared by dissolving 15.0 g of NHs in 250.0 g of water. What is the molality of the NHs in the solution? Molar mass/g mo NH3 17.03 (A) 0.00353 m (C) 3.52 nm (B) 3.24 m (D) 60.0 m

Explanation / Answer

4. (C)

Heat required to vaporize 1 mole (159.81gm) Br2 = 15kJ

Heat required to vaporize 1 gm Br2= (15/159.81) kJ = 0.094 kJ

Heat required to vaporize 25 gm Br2= 0.094 x 25 = 2.35 kJ

5. (B)

12% by mass means 12 gm of sodium carbonate is preapre in 100 gm of solution.. .... ~statement 1...

105.99 gm of sodium carbonate = 1 mole

1 gm of sodium carbonate = 1/105.99 moles

12 gm of sodium carbonate = (1/105.99) × 12 = 0.1132

Now,

Density of solution = 1.124

Mass of solution/ volume of solution = 1.124 (given)

Mass of solution =100gm ( from statement 1 )

Therefore,

100/V = 1.124

V= 88.96 mL = 0.08896 L

Molar concentration = number of moles / volume of solution

= 0.1132/ 0.08896 =1.27 M

6.(C)

17.03 gm of NH3= 1 mole

1 gm of NH3= (1/17.03) moles

15 gm of NH3= (1/17.03)×15 = 0.88 moles

Mass of water = 250 gm = 0.25 kg

Molality = number of moles/ volume of solvent = 0.88/0.25= 3.523m

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