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One litre N2, 7/8 litre O2 and 1 litre CO are taken in a mixture under indentical conditions of P and T. The amount of gases present in mixture is given by:
(c) = 1xPx28/RT; = 1xPx28/RT; =7/8*Px32/RT
Equal volumes of 0.1 M AgNO3 and 0.2 M NaCl are mixed. The concentration of N ions in the mixture will be:
(b) M of AgNO3 = 0.1 x V
M of NaCl = 0.2 x V
M of N = 0.1 x V and total V= 2V [ Ag NO3 (V) = NO3-(V) , So that total volume becomes 2V]
[ N] = 0.1 xV/2V =0.05
A solution contains Na2CO3 and NaHCO3. 10 mL of the solution required 2.5 mL of 0.1 M H2SO4 for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further 2.5 mL of 0.2 M H2SO4 was required. The amount of Na2CO3 and NaHCO3 in 1 litre of the solution is:
(a) For phenolphthalein:
1/2 M eq. of Na2CO3 = 2.5x0.1x2=0.5
For methyl orange:
1/2 M eq. of Na2CO3 + M eq. of NaHCO3 = 2.5x0.2x2=1.0
M eq. of NaHCO3 = 0.5 and M eq. of Na2CO3 =1.0
w/84*1000=0.5 w/53*1000=1
w=0.042g in 10 mL w=0.053g in 10mL
w= 4.2g in 1 litre w= 5.3 g in 1 litre
A sample of pure Cu (3.18g ) heated in a stream of oxygen for some time gains in mass with the formation of black oxide of copper (CuO). The final mas is 3.92 g. What per cent of copper remains unoxidised?
(c) Let a g of Cu be oxidised to give CuO, i.e.,
Thus, final mass = =3.92
a = 2.94g
Thus, % of Cu left unoxidised = =7.55%
One mole of a mixture of CO and CO2 requires exactly 20g of NaOH in solution for complete conversion of all the CO2 into Na2CO3. How much NaOH would it require for conversion into Na2CO3, if the mixture (one mole) is completely oxidised to CO2?
The mass of 50% (w/w) solution of HCl required to react with 100g of CaCO3 would be:
(c) Equivalent of HCl = Equivalent of CaCO3
Thus, w/36.5=100/50;
w=73g HCl;
50g HCl is presenting 100g HCl solution and thus, weight of solution required for, 73 g HCl = 73*100/50=146g.
Versene, a chelating agent having chemical formula C2H4N2(C2H2O2Na)4. If each mole of this compound could bind 1 mole of Ca2+, then the rating of pure versene expressed as mg of CaCO3 bound per g of chelating agent is:
(d) 1 mole ca2+ = 1 mole CaCO3 = 100g
Rating = mg of CaCO3 needed per g chelating agent (molar mass= 380) = 100x103/380 = 263mg
A gas is found to have the formula (CO)x. Its VD is 70. The value of x must be:
(c) Molar mass = 70 x 2 =140;
(CO)x, (12+16).x =140
therefore, x=5
100 mL of PH3 when decomposed produces phosphorus and hydrogen. The change in volume is:
(a) 2PH3(g) 2P(s) + 3H2(g)
100 0 0 Before dissociation
0 - 150 After dissociation
More explanation below:
As the volume occupied by gases only depends on number of moles of gases, the answer is related to number of moles of gases produced.
Since, no temperature and pressure is given, one can assume NTP.
At NTP, the reaction produces Phosphorous(Solid) and Hydrogen(gas). Looking at the balanced equation above, 2 moles of PH3 gas would give 3 moles of H2 gas.
Therefore, there is 50% increase in number of gaseous molecules after reaction. (2 to 3 moles)
As the volume depends on number of moles, the increased volume is also 50% more than original volume.
Increase in volume = Original Volume * 50 % = 100 ml * 50% = 50ml
New Volume = Original Volume + Increase in Volume = (100 + 50) ml = 150 ml
The molality of 1M solution of NaCl ( specific gravity 1.0585 g/mL ) is:
(b) Mass of solvent = mass of solution - mass of NaCl
= 1.0585 x 1000 - 58.5
= 1058.5 - 58.5 = 1000g =1kg
m = mole of NaCl / mass of solvent in kg = 1/1 = 1
The total molarity of all the ions containing 0.1 M of CuSO4 and o.1M of Al2(SO4)3 is:
(b) Mole of Cu2+ = 0.1x1 = 0.1
Mole of S=0.1x1=0.1
Mole of Al3+ = 0.1x2=0.2
Mole of S = 0.1x3 = 0.3
Total moles of ions present in1 litre =0.7
Molarity of all ions = 0.7M
A 0.1097 gm sample of As2O3 require 26.10 ml of KMnO4 solution for its titration. The molarity of KMnO4 solution is
A sample of CaCO3 is 50% pure. On heating 1.12 litres of CO2 (at NTP) is obtained. Residues left (assuming non-volatile impurity) is , (Ca = 40, C= 12, O=16)
Solution approach
CaCO3 → CaO + CO2 (decomposition reaction for 50% pure part on heating)
The impure part doesn't react as per the question, it remains as it is. So, total residue after heating will be
1) mass of the impure part + mass of CaO (left from pure part)
since vol of CO2
suggest that moles are 1.12/22.4 = 0.05 moles
so wt of = 0.05 .100 = 5 gram. since 50 % is impure , so 5 gram is impure
You will get mass of impure part = 5g & mass of CaO = 2.8
So total = 5 + 2.8 = 7.8g
2SO2 + O2 2SO3 6.4 gm SO2 and 3.2 gm O2 to form SO3 . How much maximum mass of SO3 is formed?
0.5 mole of BaCl2 is reacted with 0.2 mole Na3PO4 then maximum moles of Ba3(PO4)2 formed is
3BaCl2 + 2Na3PO4 Ba3(PO4)2 +6NaCl
0.5mole 0.2 mole
Na3PO4 is the limiting reagent
2 Mole Na3PO4 2 Mole Ba3(PO4)2 +6NaCl
0.2 Mole Na3PO4 0.1 Mole Ba3(PO4)2