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0.5 g of fuming H2SO4 (oleum) is diluted with water. This solution is completely neutralised by 26.7 mL of 0.4 N NaOH. The percentage of free SO3 in the sample is:
(c) Meq. of H2SO4 + Meq.Of SO3 = Meq. of NaOH
(0.5-a)/49*1000 +a/40*1000 = 26.7 x 0.4
a = 0.103
% of SO3 = 0.103/0.5*100 = 20.6%
A metal M forms a compound M2HPO4. The formula of the metal suphate is:
(a) M2HPO4 means valence of metal is one and thus, sulphate of metal is M2SO4.
A partially dried clay mineral contains 8% water. The original sample contained 12% water and 45% silica. The% of silica in the partially dried sample is nearly:
(d) Silica Water Clay Mineral
45 12 43 Initial %
a 8 (92-a) % after heating
( in dry state ,8% loss of water is there ,so remaining weight is 100 - 8 = 92)
The % ratio of silica and clay remains constant on heating
i.e.,
a = 47%
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
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