Test: Quantitative Analysis (NCERT)
10 Questions MCQ Test Chemistry Class 11 | Test: Quantitative Analysis (NCERT)
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0.92 g of an organic compound was analysed by combustion method. The mass of the U-tube increased by 1.08 g. What is the percentage of hydrogen in the compound?
Increase in mass of U-tube = 1.08 g
Mass of water formed = 1.08 g
18gH20 = 2gH
An organic compound gave 0.4655 g of CO2 on complete combustion. If the mass of the compound taken was 0.2115 g, what is the percentage of C in it?
Mass of CO2 formed = 0.4655 g
Mass of organic compound taken = 0.2115 g
In Duma'smethod 0.52 g of an organic compound on combustion gave 68.6 mL N2 at 27oC and 76 mm pressure. What is the percentage of nitrogen in the compound?
V1 = 68.6 mL, P1 =. 756 mm, T1 = 300 K V2 = ?, P2 = 760 mm, T2 = 273 K
Percentage of nitrogen in organic compound
In kjeldahl's method of estimation of nitrogen, nitrogen is quantitatively converted to ammonium sulphate. It is the treated with the standard solution of alkali. The nitrogen which is present is estimated as:
When treated with standard solution of alkali nitrogen is liberated as NH3 which is absorbed in H2SO4 to get percentage of nitrogen.
In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. What is the percentage of bromine in the compound?
Molar mass of AgBr = 108 + 80 = 188 g mol-1
Percentage of Br2 = 
2.18 g of an organic compound containing sulphur produces 1.02g of BaSO4. Thepercentage of sulphur in the compound is
1.6 g of an organic compound gave 2.6 g of magnesium pyrophosphate. The percentage of phosphorus in the compound is
Match the column I with column II in which formula for estimation of an element is given and mark the appropriate choice.
(w1 = weight of CO2, w= weight of organic compound)
(V = volume of N2 at STP, w = weight of organic compound)
In D2O percentage of deuterium
% of oxygen - 100 - 20 = 80%
0.2 g of an organic compound contains C, H and O.On combustion, it yields 0.15 g CO2 and 0.12 g H2O.Ihe percentage of C, H and O respectively is
% of O = 100 - (20 + 6.66) = 73.34%
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