➡️Solve the above problems in ur notebook.....
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➡️Solve the above problems in ur notebook.....
⚠️No spam...wrna Corona virus hojaye ga!!....xd
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1) We know that:
Mass percentage of C6H6
and we also know that:
Mass percentage of CCl4
Alternatively,
Mass percentage of CCl4= (100 - 15.28)%
= 84.72%
2) Let the mass of solution be 100 g
Mass of Benzene = 30 g
Mass of carbon tetrachloride = 100 g - 30 g = 770 g
Number of moles of benzene = Mass / Molar mass
= 30 g / 78 g mol⁻¹
= 0.385 mol
Number of moles of carbon tetrachloride = Mass / Molar mass
= 70 g / 154 g mol⁻¹
= 0.455 mol
Mole fraction of benzene = Mass of benzene / Moles of benzene + Moles of tetrachloride
= 0.385 mol / 0.840 mol
FINAL RESULT :- 0.458
3) (a) weight of Co(NO3)2.6H2O = 30g
molar mass of Co(NO3)2.6H2O = 291 g/mol.
volume of solution = 4.3 L
so, number of moles of solute = given weight/molar mass
= 30/291 = 0.103 mol
molarity = 0.103mol/4.3L = 0.023M
(b) Given,
30 mL of 0.5 M H2SO4 diluted to 500 mL.
In 1000 mL of 0.5 M H2SO4, number of moles present is 0.5 mol.
∴ In 30 mL of 0.5 M H2SO4, number of moles present = 30 × 0.5/1000 = 0.015 mol
∴ molarity = number of moles present/volume
= 0.015mol/volume of solution
= 0.015/0.5L
= 0.03M
4) Mass of required aqueous solution = 2.5 kg = 2500 g
0.25 molal aqueous solution of urea it means that 0.25 mole of urea is dissolved in 1000 grams of water.
Mass of water = 1000 g
Mass of urea = 0.25 mol
Molar mass of urea (H₂NCONH₂) = 4 × 1 + 2 × 14 + 1 × 12 + 1 × 16 = 60 g mol⁻¹
NOTE :- Mass of urea = Number of moles of urea × Molar mass of urea
Mass of 0.25 moles of urea = 0.25 mol ×× 60 h mol⁻¹ = 15 g
Mass of solution = 1000 g + 15 g = 1015 g
1015 g of aqueous solution contains urea = 15 g
∴ 2500 g of aqueous solution will require urea
= 15 g / 1015 kg × 2500 g
= 36.95 g
5) Molar Mass of Kl = 166 g/mol
Density = 1.202 g/ml
Mass/Mass % = 20%
Mass of solution = 100g
Mass of solute = 20g
Mass of solvent = 80g = 0.08 kg
Density = Mass of solution / Volume of solution
1.202 = 100 / V
V = 100/1.202
V = 83.19 ml
V = 0.083 L
Moles of Kl = Given mass/Molar Mass = 20/166 = 0.12
-----------------------------------------------------
a) Molality = Moles of Kl /Mass of solvent (Kg)
= 0.12 / 0.08
= 1.5 m
-----------------------------------------------------
b) Molarity = Moles of solute / Volume of solution
= 0.12 / 0.083
= 1.44 M
------------------------------------------------------
c) Mole Fraction = Mole of solute / Moles of solute + Moles of solvent
= 0.12 / ( 0.12 + 4.44 )
= 0.12 / 4.56
= 0.026
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HOPE THIS HELPS....
@HARSHA