0

Guest

Courses New

(a) One litre of a sample of hard water contain 1 mg of CaCI 2 and 1 mg of MgCI 2 . Find the total hardness in terms of parts of CaCO 3 per 10 6 parts of water by weight. (b) A sample of hard water contains 20 mg of Ca ++ ions per litre. How many milli-equivalent of Na 2 CO 3 would be required to soften 1 litre of the sample? (c) 1 gm of Mg is burnt in a closed vessel which contains 0.5 gm of O 2 . (i) Which reactant is left in excess? (ii) Find the weight of the excess reactants. (iii) How may milliliters of 0.5 N H 2 SO 4 will dissolve the residue in the vessel.

(a) One litre of a sample of hard water contain 1 mg of CaCI2 and 1 mg of MgCI2. Find the total hardness in terms of parts of CaCO3 per 106 parts of water by weight.
(b) A sample of hard water contains 20 mg of Ca++ ions per litre. How many milli-equivalent of Na2CO3 would be required to soften 1 litre of the sample?
(c) 1 gm of Mg is burnt in a closed vessel which contains 0.5 gm of O2.
(i) Which reactant is left in excess?
(ii) Find the weight of the excess reactants.
(iii) How may milliliters of 0.5 N H2SO4 will dissolve the residue in the vessel.

Grade:10

4 Answers

Deepak Patra
askIITians Faculty 471 Points
9 years ago
. E = 12400 / λ (in Å) eV = 12400 / 975 = 12.75 eV … (i)
Also
13.6 [ 1 / n21 – 1 / n22] = 12. 75 ⇒ [1 /1 – 1 / n22] = 12.75 / 13. 6 ⇒ n2 = 4
For every possible transition one downward arrow is shown therefore the possibilities are 6.
Note : For longest wavelength, the frequency should be smallest.
This corresponds to the transition from n = 4 to n = 3, the energy will be E4 = 13.6 / 42 ; E3 = - 13.6 / 32
∴ E4 – E3 = 13.6 / 42 – (- 13.6 / 32) = 13.6 [1/9 – 1/16]
= 0.66 eV = 0.66 x 1.6 x 10-19 J = 1.056 x 10-19 J
Now, E = 12400 / λ (inÅ) eV ∴ λ = 18787 Å
Jitender Pal
askIITians Faculty 365 Points
9 years ago
(a) CaCI2 ≡ CaCO3 ≡ MgCI2
M. wt. 111 100 95
From this it is evident, the
111 mg CaCI2 will give CaCO3 = 100mg
∴ 1 mg CaCI2 will give CaCO3 = 100/111 mg = 0.90 mg
95 mg MgCI2 gives CaCO3 = 100mg
∴ 1 mg MgCI2 gives CaCO3 = 100/95 mg = 1.05 mg
∴ Total CaCO3 formed by 1 mg CaCI2 and 1 mg MgCI2 = 0.90 + 1.05 = 1.95 mg
∴ Amount of CaCO3 present per litre of water = 1.95mg
∴ wt of 1 ml of water = 1g = 103 mg
∴ wt of 100 ml of water = 103 * 103 = 106mg
∴ Total hardness of water in terms of parts of CaCO3 per 106 parts of water by weight = 1.95 parts.
(b) Eqwt of Ca++ = Mol.wt/Charge = 40/2 = 20
Ca2+ + Na2CO3 → CaCO3 + 2Na+
1 milliequivalent of Ca2+ = 20 mg
1 milliequivalentof Na2CO3 is required to soften 1 litre of hard water.
(c) 2Mg + O2 → 2MgO
2 * 24 = 48g 32g 2(24 + 16) = 80g
∵ 32g of O2 reacts with = 448g Mg
∴ 0.5g of O2 reacts with = 48/32 * 0.5 = 0.75g
Weight of unreacted Mg = 1.00 – 0.75 = 0.25g
Thus Mg is left in excess.
Weight of MgO formed = 80/48 * 0.75 = 1.25g
MgO + H2SO4 → MgSO4 + H2O
(40g)
According to reaction
∵ 40g MgO is dissolved it gives 1000 ml of 1 N. H2SO4
∴ 40 g MgO is dissolved it gives 2000 ml 0.5 N H2SO4
∴ 1.25 MgO is dissolved it gives
= 2000 *1.25/40 ml of 0.5 N H2SO4
= 62.5ml of 0.5N H2SO4
Ifjdndbishevehe
11 Points
5 years ago
Hhhh should we just hang in the morning before you guys leave the school area in a little while I don’t want you going I have a lot to talk about tomorrow I have a good weekend thanks for your nice day and I hope you’re doing better thanks bye hello bye hi hello hi bye hello hello hi hello hello happy birthday Get low with the whistle blow 
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

(a) CaCI2 ≡ CaCO3 ≡ MgCI2
M. wt. 111 100 95
From this it is evident, the
111 mg CaCI2 will give CaCO3 = 100mg
∴ 1 mg CaCI2 will give CaCO3 = 100/111 mg = 0.90 mg
95 mg MgCI2 gives CaCO3 = 100mg
∴ 1 mg MgCI2 gives CaCO3 = 100/95 mg = 1.05 mg
∴ Total CaCO3 formed by 1 mg CaCI2 and 1 mg MgCI2 = 0.90 + 1.05 = 1.95 mg
∴ Amount of CaCO3 present per litre of water = 1.95mg
∴ wt of 1 ml of water = 1g = 103 mg
∴ wt of 100 ml of water = 103 * 103 = 106mg
∴ Total hardness of water in terms of parts of CaCO3 per 106 parts of water by weight = 1.95 parts.
(b) Eqwt of Ca++ = Mol.wt/Charge = 40/2 = 20
Ca2+ + Na2CO3 → CaCO3 + 2Na+
1 milliequivalent of Ca2+ = 20 mg
1 milliequivalentof Na2CO3 is required to soften 1 litre of hard water.
(c) 2Mg + O2 → 2MgO
2 * 24 = 48g 32g 2(24 + 16) = 80g
∵ 32g of O2 reacts with = 448g Mg
∴ 0.5g of O2 reacts with = 48/32 * 0.5 = 0.75g
Weight of unreacted Mg = 1.00 – 0.75 = 0.25g
Thus Mg is left in excess.
Weight of MgO formed = 80/48 * 0.75 = 1.25g
MgO + H2SO4 → MgSO4 + H2O
(40g)
According to reaction
∵ 40g MgO is dissolved it gives 1000 ml of 1 N. H2SO4
∴ 40 g MgO is dissolved it gives 2000 ml 0.5 N H2SO4
∴ 1.25 MgO is dissolved it gives
= 2000 *1.25/40 ml of 0.5 N H2SO4
= 62.5ml of 0.5N H2SO4

Thanks and Regards

Think You Can Provide A Better Answer ?

Other Related Questions on Physical Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More