1. Xtra Edge February 2012 - Career Point
1. Xtra Edge February 2012 - Career Point 1. Xtra Edge February 2012 - Career Point
Sol. Mass of sample of feldspar containing Na2O and K2O = 0.5 g. According to the question, Na2O + 2HCl → 2NaCl + H2O ..(1) 2 × 23 + 16 = 62g 2(23 + 35.5) = 117 g K2O + 2HCl → 2KCl + H2O ...(2) 2 × 39 + 16 = 94g 2(39 + 35.5) = 149 g Mass of chlorides = 0.1180 g Let, Mass of NaCl = x g ∴ Mass of KCl = (0.1180 – x)g Again, on reaction with silver nitrate, NaCl + AgNO3 → AgCl + NaNO3 ...(3) 23 + 35.5 = 58.5g 108 + 35.5 = 143.5g KCl + AgNO3 → AgCl + KNO3 ...(4) 39 + 35.5 = 74.5g 108 + 35.5 = 143.5g Total mass of AgCl obtained = 0.2451 g Step 1. From eq. (3) 58.5 g of NaCl yields = 143.5 g AgCl 143. 5 ∴ x g of NaCl yields = x g AgCl 58. 5 And from eq. (4), 74.5 g of KCl yields = 143.5 g of AgCl ∴ (0.1180 – x)g of KCl yields 143. 5 = (0.1180 – x)g AgCl 74. 5 Total mass of AgCl 143. 5 58. 5 x + 143. 5 74. 5 (0.1180 – x) = 0.2451 which gives, x = 0.0342 Hence, Mass of NaCl = x = 0.0342 g And Mass of KCl = 0.1180 – 0.0342 = 0.0838g Step 2. From eq.(1), 117 g of NaCl is obtained from = 62 g Na2O ∴ 0.0342 g NaCl is obtained from 62 = × 0.032 = 0.018 g Na2O 117 From eq. (2), 149 g of KCl is obtained from = 94 g K2O ∴ 0.0838 g of KCl is obtained from 94 = × 0.0838 = 0.053 g K2O 149 Step 3. % of Na2O in feldspar = % of K2O in feldspar = 0. 053 0. 5 0. 018 0. 5 = 3.6% × 100 × 100 = 10.6 % 8. 5 ml of 8 N of nitric acid, 4.8 ml of 5 N hydrochloric acid and a certain volume of 17 M sulphuric acid are mixed together and made upto 2 litre. 30 ml of this mixture exactly neutralizes 42.9 ml of sodium carbonate solution containing one gram of Na2CO3.10H2O in 100 ml of water. Calculate the amount in grams of the sulphate ions in solution. Sol. Given that, [IIT-1985] N HNO = 8 N 3 V HNO = 5 ml 3 NHCl = 5 N VHCl = 4.8 ml M H = 17 M 2SO4 Step 1. Meq. of HNO3 in 2L solution = N HNO × 3 V HNO3 = 8 × 5 = 40 ∴ Meq. of HNO3 in 30 ml solution 40 = × 30 = 0.6 2000 Step 2. Meq. of HCl in 2L solution = NHCl × VHCl = 5 × 4.8 = 24 ∴ Meq. of HCl in 30 ml solution 24 = × 30 = 0.36 2000 Step 3. Meq. of H2SO4 in 2L solution = Valency factor × M × = 2 × 17 × V H2SO4 XtraEdge for IIT-JEE 10 FEBRUARY 2012 H2SO 4 V H2SO4 ∴ Meq. of H2SO4 in 30 ml solution 34× VH × 30 2SO4 = = 0.51 V H2SO4 2000 Step 4. Also given that, Volume of Na2CO3.10H2O = 100 ml Mass of Na2CO3.10H2O = 1 g Equivalent mass of Na2CO3.10H2O Molecular mass 286 –1 = = = 143 g equiv 2 2 We know, Normality Mass of solute × 1000 = Equivalent mass of solute × Volume (ml) 1× 1000 = = 0.070 N 143× 100 Meq. of Na2CO3.10H2O = Na2CO3. 10H2O V = 0.070 × 42.9 = 3.003 N × Na 2CO3. 10H2O
Step 5. 1 gram equivalent acid neutralizes 1 gram equivalent of base. ∴ 0.6 + 0.36 + 0.51 V H = 3.003 2SO4 Solving , V H = 4 ml 2SO4 Step 6. 1000 ml of 1 M H2SO4 contains = 96 g SO4 2– ions 96 × 17× 4 ∴ 4ml of 17 M H2SO4 contains = 1000 = 6.528g SO4 2– ions 9. The equilibrium constant Kp of the reaction 2SO2(g) + O2(g) 2SO3(g) is 900 atm –1 at 800 K. A mixture containing SO3 and O2 having initial partial pressures of 1 atm and 2 atm, respectively, is heated at constant volume to equilibrate. Calculate the pressure of each gas at 800 K. [IIT- 1989] Sol. Since to start with SO2 is not present, it is expected that some of SO3 will decompose to give SO2 and O2 at equilibrium. If 2x is the partial pressure of SO3 that is decreased at equilibrium, we would have 2SO2(g) + O2(g) 2SO3(g) t = 0 0 2 atm 1 atm teq 2x 2 atm + x 1 atm – 2x 2 ( p ) 2 SO3 ( 1 atm − 2x) Hence, Kp = = 2 2 ( pSO ) ( p ) 2 O ( 2x) ( 2 atm + x) 2 = 900 atm –1 Assuming x
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Sol. Mass of sample of feldspar containing Na2O and<br />
K2O = 0.5 g.<br />
According to the question,<br />
Na2O + 2HCl → 2NaCl + H2O ..(1)<br />
2 × 23 + 16 = 62g 2(23 + 35.5) = 117 g<br />
K2O + 2HCl → 2KCl + H2O ...(2)<br />
2 × 39 + 16 = 94g 2(39 + 35.5) = 149 g<br />
Mass of chlorides = 0.1180 g<br />
Let, Mass of NaCl = x g<br />
∴ Mass of KCl = (0.1180 – x)g<br />
Again, on reaction with silver nitrate,<br />
NaCl + AgNO3 → AgCl + NaNO3 ...(3)<br />
23 + 35.5 = 58.5g 108 + 35.5 = 143.5g<br />
KCl + AgNO3 → AgCl + KNO3 ...(4)<br />
39 + 35.5 = 74.5g 108 + 35.5 = 143.5g<br />
Total mass of AgCl obtained = 0.2451 g<br />
Step <strong>1.</strong> From eq. (3)<br />
58.5 g of NaCl yields = 143.5 g AgCl<br />
143.<br />
5<br />
∴ x g of NaCl yields = x g AgCl<br />
58.<br />
5<br />
And from eq. (4),<br />
74.5 g of KCl yields = 143.5 g of AgCl<br />
∴ (0.1180 – x)g of KCl yields<br />
143.<br />
5<br />
= (0.1180 – x)g AgCl<br />
74.<br />
5<br />
Total mass of AgCl<br />
143.<br />
5<br />
58.<br />
5<br />
x +<br />
143.<br />
5<br />
74.<br />
5<br />
(0.1180 – x) = 0.2451<br />
which gives, x = 0.0342<br />
Hence, Mass of NaCl = x = 0.0342 g<br />
And Mass of KCl = 0.1180 – 0.0342 = 0.0838g<br />
Step 2. From eq.(1),<br />
117 g of NaCl is obtained from = 62 g Na2O<br />
∴ 0.0342 g NaCl is obtained from<br />
62<br />
= × 0.032 = 0.018 g Na2O<br />
117<br />
From eq. (2),<br />
149 g of KCl is obtained from = 94 g K2O<br />
∴ 0.0838 g of KCl is obtained from<br />
94<br />
= × 0.0838 = 0.053 g K2O<br />
149<br />
Step 3. % of Na2O in feldspar =<br />
% of K2O in feldspar =<br />
0.<br />
053<br />
0.<br />
5<br />
0.<br />
018<br />
0.<br />
5<br />
= 3.6%<br />
× 100<br />
× 100 = 10.6 %<br />
8. 5 ml of 8 N of nitric acid, 4.8 ml of 5 N<br />
hydrochloric acid and a certain volume of 17 M<br />
sulphuric acid are mixed together and made upto 2<br />
litre. 30 ml of this mixture exactly neutralizes 42.9<br />
ml of sodium carbonate solution containing one<br />
gram of Na2CO3.10H2O in 100 ml of water.<br />
Calculate the amount in grams of the sulphate ions<br />
in solution.<br />
Sol. Given that,<br />
[IIT-1985]<br />
N HNO = 8 N<br />
3<br />
V HNO = 5 ml<br />
3<br />
NHCl = 5 N<br />
VHCl = 4.8 ml<br />
M H = 17 M<br />
2SO4<br />
Step <strong>1.</strong> Meq. of HNO3 in 2L solution<br />
=<br />
N HNO ×<br />
3<br />
V HNO3<br />
= 8 × 5 = 40<br />
∴ Meq. of HNO3 in 30 ml solution<br />
40<br />
= × 30 = 0.6<br />
2000<br />
Step 2. Meq. of HCl in 2L solution<br />
= NHCl × VHCl<br />
= 5 × 4.8 = 24<br />
∴ Meq. of HCl in 30 ml solution<br />
24<br />
= × 30 = 0.36<br />
2000<br />
Step 3. Meq. of H2SO4 in 2L solution<br />
= Valency factor × M ×<br />
= 2 × 17 ×<br />
V H2SO4<br />
<strong>Xtra</strong><strong>Edge</strong> for IIT-JEE 10 FEBRUARY <strong>2012</strong><br />
H2SO<br />
4<br />
V H2SO4<br />
∴ Meq. of H2SO4 in 30 ml solution<br />
34×<br />
VH<br />
× 30<br />
2SO4<br />
=<br />
= 0.51 V H2SO4<br />
2000<br />
Step 4. Also given that,<br />
Volume of Na2CO3.10H2O = 100 ml<br />
Mass of Na2CO3.10H2O = 1 g<br />
Equivalent mass of Na2CO3.10H2O<br />
Molecular mass 286 –1<br />
=<br />
= = 143 g equiv<br />
2 2<br />
We know,<br />
Normality<br />
Mass of solute × 1000<br />
=<br />
Equivalent mass of solute × Volume (ml)<br />
1×<br />
1000<br />
= = 0.070 N<br />
143×<br />
100<br />
Meq. of Na2CO3.10H2O<br />
= Na2CO3. 10H2O<br />
V<br />
= 0.070 × 42.9 = 3.003<br />
N × Na 2CO3.<br />
10H2O
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